Intermediate Classical Mechanics - University of Floridathorn/homepage/cmuglectures.pdf · The subject of mechanics deals with the motion ... for relativistic mechanics and even in

Intermediate Classical Mechanics

Charles B. Thorn1

Institute for Fundamental Theory

Department of Physics, University of Florida, Gainesville FL 32611

Abstract

1E-mail address: [email protected]

1 c©2013 by Charles Thorn

Contents

1 Introduction 41.1 Newton’s Three Laws of motion . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Vector notation, algebra, and calculus 52.1 Vector notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Addition of two vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.3 Scalar product of two vectors . . . . . . . . . . . . . . . . . . . . . . . . . . 62.4 Vector product of two vectors . . . . . . . . . . . . . . . . . . . . . . . . . . 62.5 Some force laws in vector notation . . . . . . . . . . . . . . . . . . . . . . . . 72.6 Differentiating vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.7 Momentum Conservation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.8 Inertial Frames of Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.9 Coordinate Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.10 Polar coordinates in a plane . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.11 Newtonian Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.12 The Value of New Formulations . . . . . . . . . . . . . . . . . . . . . . . . . 12

3 Simple motions in one and two dimensions 133.1 Projectile motion without air resistance . . . . . . . . . . . . . . . . . . . . . 133.2 Air resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.3 Linear air resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.4 Millikan oil drop measurement of electric charge. . . . . . . . . . . . . . . . . 153.5 Quadratic air resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163.6 A charged particle in a uniform magnetic field . . . . . . . . . . . . . . . . . 18

4 Momentum and Angular Momentum 204.1 Momentum Conservation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204.2 Rockets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204.3 The Center of Mass Coordinate . . . . . . . . . . . . . . . . . . . . . . . . . 214.4 Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224.5 Systems of several particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

5 Energy 245.1 Conservative forces and potential energy . . . . . . . . . . . . . . . . . . . . 245.2 Stoke’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255.3 Potential energy examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265.4 One dimensional systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275.5 Central forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295.6 Energy of systems of particles . . . . . . . . . . . . . . . . . . . . . . . . . . 32


6 Oscillations 356.1 Descriptions of Simple Harmonic Motion . . . . . . . . . . . . . . . . . . . . 366.2 2 and 3 dimensional oscillations . . . . . . . . . . . . . . . . . . . . . . . . . 376.3 Damped Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386.4 Driving oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 396.5 Resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416.6 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 426.7 RMS Displacement, Parseval’s Theorem . . . . . . . . . . . . . . . . . . . . 43

7 Variational Principles 447.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447.2 Calculus of Variations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 457.3 Nonmechanics applications of the calculus of variations. . . . . . . . . . . . . 46

8 Hamilton’s Principle of Least (Stationary) Action 488.1 Generalized coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 488.2 The Action and Hamilton’s Principle . . . . . . . . . . . . . . . . . . . . . . 488.3 Changing Coordinates in the Lagrangian . . . . . . . . . . . . . . . . . . . . 498.4 The energy from the Lagrangian . . . . . . . . . . . . . . . . . . . . . . . . . 508.5 The simple pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 518.6 Constraints in general . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 518.7 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53


1 Introduction

The subject of mechanics deals with the motion of material bodies in three dimensionalspace. The bodies might be fundamental particles like the electron, or extended objectslike baseballs, planets, or stars. We usually distinguish material bodies from other physicalentities such as electromagnetic or gravitational fields, whose description goes beyond themechanics we study in this course.

The subject of this course is classical mechanics which neglects quantum effects. Inother words we pretend that Planck’s constant ~ = h/2π is zero. This is a very goodapproximation for all of the everyday physics we encounter around us. Quantum effectsbecome important only when we delve into atomic and molecular scales. For the most part,in this first semester, we also assume that all speeds are much smaller than c the speed oflight, so that the modifications of mechanics required by Einstein’s theory of relativity canalso be neglected. Relativistic physics will be deferred to the second semester of the course.

Thus in this first semester we will be studying the application of Newton’s laws of motion,which have been in place since the seventeenth century.

1.1 Newton’s Three Laws of motion

We start with an idealization: we imagine that we can isolate a material body from allexternal influences (we say that no forces act on the body). In the real physical worldthis idealization is never exactly realized but it can be approached by gradually reducingfrictional forces and for example restricting motion near the earth’s surface to a horizontalplane. Newton’s first law then describes the body’s motion in this idealized situation:

1st Law: In the absence of forces a body moves with uniform velocity.

It is important to note that the velocity is a vector quantity having a direction as well asmagnitude (speed). Thus the isolated body is either at rest or moves in a straight line atconstant speed. The first law is the least specific of the three laws, and in fact is validfor relativistic mechanics and even in quantum mechanics, to the extent allowed by theuncertainty principle (i.e. the isolated particle’s wave function is a plane wave!).

The second law begins to pin down the effect of forces on a body:

2nd Law: ma = F .

Here a is the vector acceleration of the body, which has a clear meaning. The mass m isan attribute of the body that quantifies its resistance to the action of the force. A heavierbody accelerates less than a lighter one if subjected to the same force. The force is notan attribute of the body, but quantifies the interaction of the body with its environment.Although not explicit in the second law there is the promise that the force is a cogent wayof characterizing the body’s interaction with its environment.

The third law puts an important constraint on the nature of forces.


3rd Law: Every action has an equal and opposite reaction.

Quantitatively it means that if body 1 exerts a force F on body 2, then of necessity body 2exerts a force −F on body 1. It is important to appreciate that Newton’s formulation of theconstraint is specific to Newtonian mechanics which includes forces which act instantaneouslyat a distance. Indeed such forces violate the causality of special relativity which requires thatinfluences can not travel faster than the speed of light. As we shall see, the third law impliesthe conservation of momentum which will hold in general, if all the momentum is accountedfor. For example, the electromagnetic field can carry momentum, and two charged bodiescan give up some of their momentum to the field. Thus the third law applied to the twobodies would fail but momentum would still be conserved.

2 Vector notation, algebra, and calculus

One of the objectives of this intermediate course on mechanics is to get beyond the over-simplifications that were made in the introductory course (PHY2048). We want to be able todescribe more complicated motions that involve all three dimensions of space. The conceptof vectors is a valuable tool for this task.

The motion of a point mass can be specified by giving its location at each time t. Todo this we establish a Cartesian coordinate system with three perpendicular coordinate axesintersecting at the origin O of the system. Any point in space can be located by its threecoordinates x, y, z. The trajectory of a point particle is then specified by the three functionsx(t), y(t), z(t).

2.1 Vector notation

Next we introduce vector notation. We are familiar with geometric notion of a vector as anarrow that points in the direction of the vector and whose length is the magnitude of thevector. We denote a vector in the text by a bold face v. On the blackboard I will frequentlyuse instead an over arrow ~v. We can multiply vectors by numbers cv. Multiplying by anegative number reverses the direction of the vector.

2.2 Addition of two vectors

The sum of two vectors v1 + v2 is given geometrically by the parallelogram rule. But amore efficient procedure, especially when several vectors are to be added, is to resolve thevector into its components along the coordinate axes. To do this we introduce unit vectorsi = x, j = y, k = z, pointing in the positive x, y, and z directions respectively. Then thecomponents of v are such that

v ≡ vxi + vyj + vzk = (vx, vy, vz) (1)

Using component notation we then have for the sum v1 +v2 = (v1x + v2x, v1y + v2y, v1z + v2z)


With this notation we can now think of the coordinates of a point in space as the position

vector:

r ≡ xi + yj + zk = (x, y, z) (2)

The position vector differs from a true vector in that it depends on the origin of coordinates.But the notation is nonetheless extremely useful. Examples of true vectors are the velocityand acceleration of a particle as well as the force. Thus Newton’s second law F = ma is anequation between two true vectors. If we have two points specified by position vectors r1

and r2 the displacement d12 = r2 − r1 is a true vector, since it doesn’t depend on the originof the coordinates system.

2.3 Scalar product of two vectors

Geometrically the scalar product of two vectors A,B is defined as A · B = |A||B| cos θwhere θ is the angle between the two vectors. In terms of components it is given by

A · B = AxBx + AyBy + AzBz =3∑

i=1

AiBi = AiBi (3)

where the last equality invokes the summation convention that repeated indices are alwayssummed. In the summation we identify x → 1, y → 2, z → 3. Along with this labeling, itis also convenient to similarly label the basis vectors:

i = (1, 0, 0) ≡ e1, j = (0, 1, 0) ≡ e2, i = (0, 0, 1) ≡ e3 (4)

ej · ek = δkj ≡{

0 j 6= k

1 j = k(5)

Then we can write any vector

A =∑

i

Aiei = Aiei, Ai = ei · A (6)

Notice that the scalar product is distributive (A + B) · C = A · C + B · C. Note also thatA · A = |A|2.

2.4 Vector product of two vectors

We have seen that the scalar product of two vectors is a number. In contrast the vectorproduct of two vectors A,B is a vector. Its magnitude is |A||B| sin θ, and its direction isperpendicular to both A and B, in the sense of the right hand rule. In terms of componentswe have

(A × B)z = AxBy − AyBx, (A × B)y = AzBx − AxBz, (A × B)x = AyBz − AzBy


By introducing the antisymmetric symbol ǫijk, the relation of components can be moresuccinctly written

(A × B)i = ǫijkAjBk (7)

where the summation convention is in force. ǫijk is antisymmetric under the interchange ofany pair of indices: ǫijk = −ǫjik = −ǫikj = −ǫkji, and therefore vanishes unless i, j, k are alldifferent. By convention ǫ123 = +1. Using ǫijk we see that

A · (B × C) = ǫijkAiBjCk = ǫkijAiBjCk = (A × B) · C (8)

The fundamental identity

ǫijkǫklm = δilδjm − δimδjl (9)

is easily proven: just put (i, j) = (1, 2), (2, 3), (3, 1) in turn. This identity implies the triplevector product identity

[A × (B × C)]i = ǫijkAj(B × C)k = ǫijkAjǫklmBlCm = (δilδjm − δimδjl)AjBlCm

= AjBiCj − A · BCi = A · CBi − A · BCi (10)

where the last form is true if B and C commute.

2.5 Some force laws in vector notation

Newton’s inverse square force law for gravitational systems can be written:

F Gravk = −mk

∑

j 6=k

Gmj(rk − rj)

|rk − rj|3(11)

where we have several particles present. The vector notation makes Newton’s third lawevident:

F kj = −Gmkmj(rk − rj)

|rk − rj|3= −F jk (12)

which has no dependence on the velocities of the particles. Newton didn’t know aboutmagnetic forces, which were understood much later. A particle of charge q moving in anelectromagnetic field experiences a force

F EM = q [E(r, t) + v × B(r, t)] (13)

The existence of magnetic forces means that we have to allow for velocity dependent forcesto hope to describe all phenomena. Note that the 3rd law has no obvious interpretation inthis formula since the fields which exert the force on the charged particle are not themselvesmaterial bodies. When the sources of the fields are included in the dynamics, however,momentum conservation will be valid with an appropriate identification of the momentumcarried by the fields.


2.6 Differentiating vectors

When we describe the trajectory of a particle as a time dependent position vector r(t), wecan obtain the velocity and acceleration of the particle as first and second derivatives:

v = r =dr

dt, a = v =

dv

dt= r =

d2r

dt2(14)

where we have introduced the dot notation for time derivatives.The time derivative of a vector is defined in parallel to the time derivative of a scalar

function:

dA

dt≡ lim

ǫ→0

A(t + ǫ) − A(t)

ǫ(15)

Since the Cartesian unit vectors i, j,k are independent of time, it is cleat that the compo-nents of dA/dt are simply (Ax, Ay, Az). In other words the Cartesian components of thetime derivative of a vector are simply the time derivatives of the Cartesian components ofthe vector.

Spatial derivatives of vector functions of the coordinates can have several manifestations.The gradient operator ∇ is defined by

∇f ≡(

∂f

∂x,∂f

∂y,∂f

∂z

)

(16)

For example the divergence and curl of a vector function A(r) are given by

∇ · A =∑

i

∂Ai

∂xi

=∂Ai

∂xi

, (∇× A)i = ǫijk∂Ak

∂xj

(17)

where the summation convention has been used. Notice that the meaning of the dot andcross products here are exactly those already defined.

For completeness we mention the integral theorems of Gauss and Stokes:∫

dV ∇ · E =

∮

dS · E,

∫

dS · ∇ × E =

∮

dl · E (18)

familiar from electromagnetic theory.

2.7 Momentum Conservation

We have mentioned that momentum conservation is a consequence of Newton’s third law.Let us first consider two particles. According to the 3rd law if particle 2 exerts a force F 12

on particle 1, then particle 1 exerts a force F 21 = −F 12 on particle 2. Then the second lawreads

dp1

dt= F 12,

dp2

dt= F 21 = −F 12 (19)

d

dt(p1 + p2) = F 12 + F 21 = 0 (20)


That is, P = p1 + p2 is a constant. If we have more particles exerting pairwise forces oneach other, the second law reads for particle k

dpk

dt=

∑

l 6=k

F kl (21)

Summing over all particles gives

dP

dt=

d

dt

∑

k

pk =∑

k

∑

l 6=k

F kl =∑

k<l

F kl +∑

k>l

F kl =∑

k<l

F kl +∑

l>k

F lk = 0 (22)

by the third law.

2.8 Inertial Frames of Reference

Inertial frames are those in which the first law holds. By implication this implicitly carriesthe assumption that we know all of the forces. If all the known forces are zero, but anunknown force is not zero we would judge the frame to be non-inertial. Once we have foundone inertial frame, any frame of reference moving at constant velocity relative to it will alsobe an inertial frame.

2.9 Coordinate Systems

We have used vector notation to express the second law concisely p = F . But to dealwith the equations mathematically it is usually best to write out the components of thevector equation. In Cartesian coordinates, the basis vectors are fixed once and for all sothat applying time derivatives to the position vector r = (x, y, z) simply means applying thederivatives to the components: r = (x, y, z). For the case where F is a constant vector thedifferential equation of motion is solved by two integrations. A familiar example is gravitynear the surface of the earth, which exerts a force −mgz on a particle of mass m. Then theforce only enters the z component:

z = −g, z = −gt + v0, z(t) = −1

2gt2 + vz

0t + z0 (23)

meanwhile x(t) = vx0 t+x0 and y(t) = vy

0t+ y0. In these solutions the 6 integration constants(vx

0 , vy0 , v

z0), (x0, y0, z0) are given by the initial conditions.

A more complicated uniform force problem is the classic freshman physics problem ofa block sliding down an inclined plane in the presence of friction. In this problem it isconvenient to pick the x-axis of the Cartesian system parallel to the inclined plane, somotion is only in the x coordinate. Then the three forces are gravity that points verticallydown mg(sin θ,− cos θ, 0), the normal force N(0, 1, 0) exerted by the plane, and the frictionalforce (−µN, 0, 0). Since there is no motion in the y coordinate, Fy = N − mg cos θ = 0.Then Fx = mg sin θ−µN = mg(sin θ−µ cos θ) determines the acceleration down the inclinedplane. then x(t) is given by

x(t) =g

2(sin θ − µ cos θ)t2 + v0t + x0 (24)


2.10 Polar coordinates in a plane

For something a little different let’s consider how the equations look in polar coordinates onthe xy-plane (z = 0).

x = r cos ϕ, y = r sin ϕ, r = (r cos ϕ, r sin ϕ, 0) (25)

Now take derivatives of r to obtain the velocity and acceleration:

v = r = r(cos ϕ, sin ϕ, 0) + rϕ(− sin ϕ, cos ϕ, 0) (26)

a = v = r(cos ϕ, sin ϕ, 0) + (2rϕ + rϕ)(− sin ϕ, cos ϕ, 0) − rϕ2(cos ϕ, sin ϕ, 0)

= (r − rϕ2)(cos ϕ, sin ϕ, 0) + (2rϕ + rϕ)(− sin ϕ, cos ϕ, 0) (27)

Now we can recognize r = (cos ϕ, sin ϕ, 0) as the unit vector pointing in the direction of r,and ϕ as the unit vector pointing in the direction of increasing ϕ at fixed r. So we can writemore compactly:

r = rr, v = rr + rϕϕ, a = (r − rϕ2)r + (2rϕ + rϕ)ϕ (28)

The reason for these complicated formulas is that the unit basis vectors r, ϕ of the polarcoordinate system are not constant in time:

dr

dt= ϕϕ,

dϕ

dt= −ϕr (29)

As a consequence, Newton’s second law in polar coordinates is a bit complicated: writeF = Frr + Fϕϕ, and find

m(r − rϕ2) = Fr, m(2rϕ + rϕ) = Fϕ (30)

The oscillating skateboard on a half pipe in the text is essentially the same as a pendulumconsisting of a point mass attached to a massless rod of length R. The mass point feelsthe downward force of gravity mg = mg(cos ϕr − sin ϕϕ) and the normal force N = −N r.Since the rod is a constant length r = r = 0, so the second law reads:

−mRϕ2 = −N + mg cos ϕ, mRϕ = −mg sin ϕ (31)

the first equation determines N = m(Rϕ2+g cos ϕ), which is precisely the force of constraintkeeping the bob moving in a circle. The important difference between the pendulum andthe skateboard on a half pipe is that the normal force exerted by the rod can be in eitherdirection, whereas the half pipe requires N > 0. We see that if ϕ > π/2 and ϕ is smallenough, N has to be negative to keep the mass moving in a circle! The second equationdetermines the motion of the pendulum

ϕ = − g

Rsin ϕ (32)


This differential equation determines ϕ(t), however not in terms of elementary functions.when ϕ ≪ 1 the right side can be approximated sinϕ ≈ ϕ and the equation linearizes,reducing to the equation of motion for a simple harmonic oscillator:

ϕ = − g

Rϕ (33)

Since the diff eq is linear with constant coefficients we can always find solutions of the formert, because then each derivative gives a factor of r and the equation reduces to r2 = −g/Ror r = ±i

√

g/R. The angular frequency of the oscillator is ω ≡√

g/R and the generalsolution is

ϕ(t) = Ae−iωt + Beiωt = (A + B) cos ωt − i(A − B) sin ωt (34)

Since ϕ(t) is a real function, we must require B = A∗. A can be determined by initialconditions on ϕ:

ϕ(t) = ϕ(0) cos ωt +ϕ(0)

ωsin ωt. (35)

The motion is periodic with period T = 2π/ω = 2π√

R/g.What can we say if ϕ is not small? We can find an energy conservation law by multiplying

the equation by ϕ:

0 = ϕϕ +g

Rϕ sin ϕ =

d

dt

(

1

2ϕ2 − g

Rcos ϕ

)

(36)

1

2ϕ2 − g

Rcos ϕ = C (37)

which we can solve for ϕ:

dϕ

dt=

√

2C + 2ω2 cos ϕ, dt =dϕ

√

2C + 2ω2 cos ϕ

t =

∫ ϕ(t)

ϕ(0)

dϕ′

√

2C + 2ω2 cos ϕ′(38)

The integral on the right is known as an elliptic integral which can not be expressed interms of elementary functions. We will return to its study later on when we come to a moresystematic discussion of energy. For small oscillations the integral can be done

t ≈∫ ϕ(t)

0

dϕ′

√

2C + 2ω2 − ω2ϕ′2=

1

ωarcsin

ωϕ(t)√2C + 2ω2

(39)

recovering simple harmonic oscillations.


2.11 Newtonian Dynamics

Classical mechanics has not really changed, in substance, since the days of Isaac Newton.The essence of Newton’s insight, encoded in his second law F = ma, is that the motion ofa particle r(t) is determined once its initial position and velocity are known. His famousequation relates the acceleration d2r/dt2 to the force on the particle, which is implicitlyassumed to depend only on the positions, and possible the velocities of the particles in thesystem. Consider a system of N particles, whose trajectories are described by 3N coordinatesrk(t), k = 1, . . . , N . Then Newton’s laws of motion take the mathematical form of 3N secondorder differential equations in time:

mkd2rk

dt2= F k({ri}, {ri}) (40)

This is the general framework, but of course for each dynamical system we also need to knowthe force law. Newton also didn’t know about Einstein’s relativity, but his law of motion,appropriately modified carries over to this domain as well. The necessary modification is toreplace ma with dp/dt, where p is the relativistic momentum of the particle:

p ≡ mv√

1 − v2/c2(41)

We see that for particles moving slowly compared to the speed of light v ≪ c, the momentumgoes approximately to its Newtonian expression p ≈ mv, so p ≈ ma. But even for fullyrelativistic motion the basic structure of Newtonian dynamics holds.

2.12 The Value of New Formulations

The previous subsection contains everything you need to know about Newtonian dynamics–once you solve the equations there is nothing more to say. However, there are other waysto look at the dynamics that reveal features of the motion that brute force solution of theequations might leave obscure. For example in elementary physics we learn how to exploitenergy conservation when the force is the gradient of a potential F = −∇V . Then insteadof working with the second order differential equations we can use energy conservation

E =1

2mv2 + V (r) = Constant (42)

to immediately read off the speed of the particle in terms of its location and total energy E.In one dimensional motion the potential energy curve tells us a lot about the types of

motion that will occur. A horizontal line of height E intersects V (x) at the “turning points”where the particle comes to rest. Points where dV/dx = 0 tell where a particle feels zeroforce. A particle placed there at rest will stay at rest: we spot the static solutions by lookingfor the maxima and minima of the potential. A minimum is stable equilibrium, whereas amaximum is unstable equilibrium.


For motion in one dimension, energy conservation implies Newton’s equations

dE

dt= mxx + x

∂V

∂x= x

(

mx +∂V

∂x

)

= 0 (43)

so as long as x 6= 0, Newton’s equation holds. But in two or more dimensions energyconservation only tells us

x · (mx + ∇V ) = 0 (44)

which only implies that mx + ∇V is perpendicular to x. For example the magnetic forcemight or might not be present:

mx = −∇V + qx × B (45)

In the second semester of this course the ideas of Lagrange, Hamilton, and Jacobi will beused to interpret general nonstatic solutions in terms of maxima or minima of an energy-likequantity called the action. Since a nonstatic solution is a curve in space rather than simplya point, we have to study the action as a function of curves, which requires the concepts ofthe calculus of variations, which we will develop near the end of this semester.

A great advantage of the action is that by construction it is an invariant under thesymmetries of the dynamical system. It is a scalar functional of the coordinates qk(t) andvelocities qk(t). It therefore summarizes the dynamical content of a system in a compactand transparent way. As we shall see, it also greatly simplifies the problem of imposingconstraints on the dynamical variables.

3 Simple motions in one and two dimensions

To gain some experience with solving Newton’s equations we go beyond uniform forces byconsidering velocity dependent forces that are independent of position. In such situations,Newton’s laws are first order differential equations for the velocity, which makes them rel-atively easy to solve. Examples of velocity dependent forces are air resistance in projectilemotion and magnetic forces in a uniform magnetic field.

3.1 Projectile motion without air resistance

When we ignore air resistance, projectile motion near the earth’s surface is governed purelyby a uniform gravitational force −mgz. The trajectory stays in the plane determined by thez-axis and the initial horizontal velocity. We choose coordinates so that this is the xz-plane.Take the initial time to be t = 0, the initial location of the particle to be x = y = 0 andz = h, and the initial velocity to be v0 = Vxx + Vzz. Then we have

x(t) = Vxt, z(t) = h + Vzt −1

2gt2 (46)

For example the range of the projectile is the horizontal distance traveled when z returns toits initial value. This happens at the time t = 2Vz/g, so the range is 2VxVz/g.


3.2 Air resistance

To add a little realism to projectile motion (and make the equations of motion a little lesstrivial!) Let’s consider adding the effect of air resistance. We will assume the force of airresistance depends only on the velocity, and is directed oppositely to the velocity (i.e. itmust tend to slow the speed!). We also make the simplifying assumption that its magnitudeis at most quadratic in v:

f ≈ −bv − cvv (47)

This should be a good approximation if v is not too large. The coefficients b, c depend on theproperties of the medium as well as the geometry of the projectile. Assuming the projectileis a sphere of diameter D, then b = βD and c = γD2, and β, γ depend on the medium. Forair at STP β = 1.6× 10−4Ns/m2 and γ = 0.25Ns2/m4. What this means for the relative sizeof the two terms is

fquad

flin

∼ 1600Dv

1m2/s(48)

which means that for normal projectiles the quadratic term dominates. This is unfortunate,because the linear term is so much easier to deal with!

3.3 Linear air resistance

The equation of motion for a projectile with linear air resistance is

mdv

dt= −mgz − bv, m

dv′

dt= −bv′, v′ = v′

0e−bt/m (49)

where we temporarily defined v′ ≡ v + (mg/b)z. Going back

v(t) = −(mg/b)z(1 − e−bt/m) + v0e−bt/m (50)

We see that in characteristic time τ = m/b, v(t) approaches the terminal velocity vter =−gτ z. It is interesting to note that the time τ and hence the terminal velocity depends onthe mass of the object, since b depends only on the shape of the object. So unlike motionin a gravitational field in vacuum, the effects of air resistance are different for varying mass.Clearly heavier particles are less affected by air resistance than lighter ones. For objects ofthe same composition (density ρ) and different size (diameter D) the mass is proportionalto D3. For linear air resistance b ∝ D, so τ ∝ D2: larger drops fall faster.

By a simple integration of v(t), we find the trajectory, with initial position at the origin(0, 0, 0),

r(t) = −(gτ)z(t − τ(1 − e−t/τ )) + v0τ(1 − e−t/τ ) (51)

x(t) = vx0τ(1 − e−t/τ ), z(t) = −(gτ)(t − τ(1 − e−t/τ )) + vz

0τ(1 − e−t/τ ) (52)


To get the shape of the trajectory, we need to eliminate t in favor of x in the expression forz:

e−t/τ = 1 − x

vx0τ

, t = −τ ln

(

1 − x

vx0τ

)

(53)

z(x) = −gτt + xvz

0 + gτ

vx0

= gτ 2 ln

(

1 − x

vx0τ

)

+ xvz

0 + gτ

vx0

(54)

Notice that z → −∞ when x → vx0τ , meaning that x will never reach that value. Thus vx

0τis an upper limit on the possible range of motion.

If air resistance is a small effect τ = m/b is large and it is reasonable to expand thelogarithm in a Taylor series:

ln(1 − a) = −a − a2

2− a3

3− · · · (55)

Putting a = x/(vx0τ), we see that the first term cancels and we get

z ≈ − gx2

2vx20

− gx3

3τvx30

+ xvz

0

vx0

= − gx

2vx20

(

x +2x2

3τvx0

− 2vx0vz

0

g

)

(56)

The range R is the value of x where the right side is 0:

R ≈ 2vx0vz

0

g− 2

3τvx0

R2 ≈ 2vx0vz

0

g− 8vx

0vz20

3τg2=

2vx0vz

0

g

(

1 − 4vz0

3vter

)

(57)

The factor in parentheses multiplies the range in vacuum, i.e. no air resistance. This factoris clearly smaller than one, but close to one in the approximation used to obtain the result.

3.4 Millikan oil drop measurement of electric charge.

The formula for the terminal velocity vter = mg/b gets modified in the presence of an electricfield to vter = (mg − qE)/b in the presence of an electric field. Thus one can turn a carefulmeasurement of the terminal velocity a charged particle into a measurement of the particle’scharge q. Looking back to the value of b = βD, with β ≈ 1.6 × 10−4, the terminal velocityof a spherical drop of oil with density ρ at zero electric field is

vter =ρV g

βD=

4π

3

D2

8β≈ π

6

9.8 · 840 × 10−12

1.6 × 10−4

m

s≈ 2.7 × 10−5 m

s(58)

for a one micron oil drop of density 840kg/m3 = 0.840g/cm3. This is so slow that it canbe directly measured by time of flight with the aid of a microscope. For example one canfine-tune the electric field to cancel the gravitational force, and then get q = −mg/E. Bymeasuring lots of different drops of the same size one learns that q is an integer multiple ofa fundamental unit of charge. Incidentally the time it takes to reach this terminal speed isτ = vver/g ≈ 3 × 10−6s, which is virtually instantaneous!.


3.5 Quadratic air resistance

Now let’s consider the case where the linear resistance force is utterly negligible, so thequadratic force must be used. Then we need to deal with the 2nd law

dv

dt= −gz − c

mvv = −gz − c

m

√

v2x + v2

y + v2zv (59)

For projectile motion we can assume vy = 0, but then we still have a hard mathematicalproblem. To see this resolve into components:

dvx

dt= − c

m

√

v2x + v2

zvx,dvz

dt= −g − c

m

√

v2x + v2

zvz (60)

We could for example use the first equation to find vz in terms of vx, vx and plug into thesecond equation, leading to a nonlinear second order differential equation!

However, if we restrict the motion to either purely horizontal or purely vertical we getsoluble equations. Let’s start with horizontal motion so gravity plays no role. So set vz =vy = 0, after which the equation for vx ≡ v becomes

dv

dt= − c

mv2,

dt

dv= − m

cv2(61)

t(v) =m

cv− m

cv0

, v(t) =mv0

cv0t + m=

v0

1 + t/τ2

(62)

where τ2 = m/(cv0), which unlike τ depends on the initial conditions. We see that the speedslows down toward 0, but only as 1/t. As it slows down, it will eventually have so low aspeed that the quadratic approximation is no longer valid, and linear air resistance takesover. After that the slowing will become exponential. By a single integration we obtain x(t),assuming the quadratic approximation:

x(t) = x0 + v0τ2 ln(1 + t/τ2) (63)

The fact that x → ∞ as t → ∞ is unrealistic and is due to the assumption that the quadraticapproximation is valid down to arbitrarily low speeds. If we put both terms into the equationof motion we get a more reasonable (though more complicated) result:

dv

dt= −bv − c

mv2,

dt

dv= − m

bv + cv2= − m

c(v + b/2c)2 − b2/4c= −m

bv+

mc/b

b + cv

t = −m

b

∫ v

v0

dv′

[

1

v′− c

b + cv′

]

= −m

bln

v(cv0 + b)

v0(cv + b), e−bt/m =

v(cv0 + b)

v0(cv + b)

v(t) =v0e

−bt/m

1 + (v0c/b)(1 − e−bt/m)∼ v0

1 + cv0t/m, for

bt

m≪ 1 (64)

x(t) = x0 +m

cln(

1 + (v0c/b)(1 − e−bt/m))

(65)

We see that after a very long time the total distance traveled x(∞)−x0 = (m/c) ln(1+v0c/b)is finite as long as b is finite.


Next we consider vertical motion, so that gravity plays a role;

mdv

dt= −mg − c|v|v, t = −m

∫ v

v0

dv′

mg + c|v′|v′(66)

Notice that we have to be careful with the sign of the resistive term. If v > 0 (upwardmotion) it should be negative so it tends to slow the projectile. But if v < 0 (downwardmotion) it must be positive to reduce the speed (magnitude of v). If v0 > 0 at the beginningof the motion both gravity and air resistance tend to slow the projectile. But then afterit reaches its high point and starts to descend, gravity and air resistance work in oppositedirections. Let us first take v0 ≤ 0, so that v will stay negative thereafter. Then the equationis solved by doing the integral

t = −m

∫ v

v0

dv′

mg − cv′2= −1

g

∫ v

v0

dv′

1 − v′2/v2ter

, vter =

√

mg

c

−gt =

∫ v

v0

dv′1

2

[

1

1 + v′/vter

+1

1 − v′/vter

]

=vter

2ln

(1 + v/vter)(1 − v0/vter)

(1 − v/vter)(1 + v0/vter)(67)

To simplify life let’s assume that v0 = 0, i.e. the ball is dropped from rest. Then invertingthe last result gives

1 + v/vter

1 − v/vter

= e−2gt/vter , v(t) = −vter1 − e−2gt/vter

1 + e−2gt/vter= −vter tanh

gt

vter

z(t) = z0 −v2

ter

gln cosh

gt

vter

= z0 +v2

ter

gln 2 − v2

ter

gln(egt/vter + e−gt/vter) (68)

∼ z0 +v2

ter

gln 2 − vtert (69)

where the last line shows the asymptotic behavior at large t, which simply reflects theapproach to a terminal velocity.

If v0 > 0, i.e. the projectile is thrown upward, the upward part of the journey is describedby

t = −m

∫ v

v0

dv′

mg + cv′2= −1

g

∫ v

v0

dv′

1 + v′2/v2ter

= −vter

garctan

v′

vter

∣

∣

∣

∣

v

v0

v(t) = vter tan

[

arctanv0

vter

− gt

vter

]

=v2

ter

g

d

dtln cos

[

arctanv0

vter

− gt

vter

]

(70)

z(t) = z(0) +v2

ter

gln cos

[

arctanv0

vter

− gt

vter

]

− v2ter

gln cos

[

arctanv0

vter

]

(71)

The upward journey ends when v(t) = 0 or gt = vter arctan(v0/vter). At that point it hasrisen an amount zmax−z(0) = (v2

ter/g) ln(√

v20 + v2

ter/vter). Then it starts falling as describedin the previous paragraph.


Finally we briefly consider general projectile motion. As we already mentioned it ishopeless to find an explicit solution to the equations of motion.

dvx

dt= − c

m

√

v2x + v2

zvx,dvz

dt= −g − c

m

√

v2x + v2

zvz (72)

We of course know what to expect qualitatively, because either type of air resistance qualita-tively hinders the motion relative to that in the vacuum. The projectile will not rise as highnor go as far as in vacuum; and of course there is a terminal velocity in both cases. Whenwe considered pure horizontal motion with quadratic air resistance we found the unreason-able result that although the velocity drops to zero, it does so slowly enough that there isno upper limit to the horizontal displacement. However this doesn’t happen in combinedvertical and horizontal motion.

To see this we look at the equations at very late times, when we know that vx → 0 andvz → vter. In this regime the equations become

dvx

dt≈ −cvter

mvx,

dvz

dt≈ −g − cvter

mvz (73)

which are the equations for linear air resistance with the parameter b = cvter. So eventuallyvx approaches zero exponentially, which means that x(t) approaches a finite value at large t:

x(t) → x∞ ≡∫ ∞

0

dtvx(t) (74)

3.6 A charged particle in a uniform magnetic field

The velocity dependent forces we have considered thus far are not fundamental: they onlytake into account the fundamental interactions of air molecules with the projectile in anaverage and over-simplified way. In contrast the magnetic force F = qv×B is a fundamentalforce of nature. Unlike the force of air resistance which dissipates energy, the magnetic forceacts in a way that conserves kinetic energy. It is fairly easy to see this2

dK

dt=

mq

22v · (v × B) = 0 (76)

because v is orthogonal to v × B! So the magnetic force is non-dissipative.Since v2 is a constant, only the direction of v changes with time. To investigate the

motion, let us choose coordinates so that B, which we shall assume is uniform, is parallelto the z axis. Then the magnetic force is parallel to the xy plane. Newton’s equation is

mdv

dt= q(v × Bz) (77)

2The complete electromagnetic force includes an electric component F = q(E + v × B). In this case wefind

dK

dt= qv · E → − d

dt(qφ) (75)

for an electrostatic field E = −∇φ. Then E = K + qφ is conserved.


Using the definition of the cross product we easily resolve this equation into components:

mdvz

dt= 0, m

dvx

dt= qBvy, m

dvy

dt= −qBvx (78)

The first equation is trivial to solve: it just says vz =constant, so z(t) = z0 + vzt. There areat least two ways to solve remaining equations.the most straightforward way is to use thesecond equation to eliminate vy from the third equation:

vy =m

qBvx, vx = −ω2vx, ω =

qB

m(79)

We recognize the second equation as the equation for trig functions cos ωt or sin ωt. Thusthe general solution for vx is

vx = C sin ωt + S cos ωt ≡ A sin(ωt − δ), vy =m

qBvx =

1

ωvx = A cos(ωt − δ) (80)

where C = A cos δ, S = −A sin δ or A =√

C2 + S2 and δ = − arctan(S/C). Then a singleintegration gives x(t), y(t):

x(t) = x0 −A

ωcos(ωt − δ), y(t) = y0 +

A

ωsin(ωt − δ) (81)

r(t) = r0 + (−R cos(ωt − δ), R sin(ωt − δ), vzt) (82)

The projection of the trajectory onto the xy plane is a circle of radius R = v0/ω = mv0/qBcentered on the point (x0, y0, 0). The particle traces this circle in a clockwise direction (Ifwe flip the sign of B the particle will move in the counter-clockwise direction). If vz 6= 0the three dimensional trajectory traces a circular helix with axis parallel to the z axis. Notethat we have 6 arbitrary constants of integration x0, y0, z0, v

z, R, δ, so that this is in fact themost general solution.

There is an interesting mathematical representation of the motion of a magnetic field interms of complex exponentials. Let us go back to the equations of motion, and define thecomplex quantity η ≡ vx + ivy. Then

dη

dt=

dvx

dt+ i

dvy

dt= ω(vy − ivx) = −iω(vx + ivy) = −iωη (83)

This is just the differential equation satisfied by e−iωt! Thus we can write the solution as

η(t) = vx(t) + ivy(t) = (v0x + iv0y)e−iωt ≡ iAeiδ−iωt, A =

√

v20x + v2

0y (84)

x(t) + iy(t) = x0 + iy0 −A

ωeiδ−iωt (85)

By taking the real and imaginary parts of both sides we recover the previous solution:

x(t) = x0 −A

ωcos(ωt − δ), y(t) = y0 +

A

ωsin(ωt − δ) (86)

We see that the complex exponential e−iωt gives a very efficient representation of uniformcircular motion. Looking down from positive z toward negative z, the motion is clockwisewhen Bz > 0 and counterclockwise when Bz < 0. This is the direction of motion thatguarantees that the magnetic force points radially toward the center of the circle.


4 Momentum and Angular Momentum

4.1 Momentum Conservation

In our discussion of Newton’s 3rd law we noted that in a closed system with no externalforces F ext = 0, the 3rd law guarantees momentum conservation:

d

dt

N∑

k=1

mkvk = F ext = 0 (87)

In a collision of two particles, momentum conservation reads

m1v1 + m2v2 = m1v′1 + m2v

′2 (88)

Suppose, for example, that particle 2 starts at rest, v2 = 0. Then the particle 2’s finalvelocity is determined:

v′2 =

m1

m2

(v1 − v′1) (89)

The description of collisions is much simpler in the inertial frame where the total momen-tum vanishes m1v1 = −m2v2. (This is called the center of mass system) Then conservationof momentum simply says that m1v

′1 = −m2v

′2. It is then very easy to calculate the change

in kinetic energy in the collision process:

∆K.E. =1

2

(

m1 +m2

1

m2

)

(

v21 − v′2

1

)

(90)

From this we immediately see that the maximum loss of kinetic energy (i.e. the collision ismaximally inelastic) occurs when v′

1 = 0 = v′2. That is when both particles in the final state

are at rest (carrying no kinetic energy). In this frame the condition for an elastic collision∆K.E. = 0 is simply that the final particles have the same speeds as the initial particles.

The Lab frame, in which particle 2 is at rest, moves relative to the CM frame with velocityV = v2CM = −(m1/m2)v1CM . In this Lab frame v1Lab = v1CM − V = (1 + m1/m2)v1CM ,v2Lab = 0, v1

′Lab = v1

′CM −V and v2

′Lab = −(m1/m2)v1

′CM −V . The condition for maximal

inelasticity, v1′CM = 0, then becomes the statement that the particles in the final state both

travel at the same velocity, −V = (m1/m2)v1CM = m1v1Lab/(m1 + m2).

4.2 Rockets

Rockets are propelled by expelling momentum in the form of its fuel exhaust, giving itselfmomentum equal to the opposite of the exhaust momentum. For simplicity consider onlymotion in a straight line. Let the particles in the exhaust have a constant velocity vex relativeto the rocket. At the moment when rocket plus fuel have mass m and velocity v, suppose


a mass −dm is expelled. Its momentum relative to an observer at rest is dm(vex − v). Thechange in the system momentum

∆Psys = (m + dm)(v + dv) − dm(v − vex) − mv = mdv + vexdm (91)

Newton’s second law gives ∆Psys = mdv + vexdm = Fextdt, but in the absence of externalforces we have simply (assuming vex is constant

dv

dm= −vex

m, v(t) − v0 = −vex ln

m(t)

m0

(92)

The smallest the final mass can be is the mass with all fuel spent. The log dependence showsthat a huge fraction of the initial mass must be fuel an discardable fuel tanks (stages) to geta substantial velocity change.

4.3 The Center of Mass Coordinate

Let’s look again at the statement of momentum conservation for an N particle system:

P =d

dt

∑

k

mkvk =d2

dt2

∑

k

mkrk = 0, M ≡∑

k

mk (93)

This equation says that the center of mass coordinate defined by R ≡ (1/M)∑

k mkrk

has zero acceleration R = 0. It therefore has time dependence R(t) = R0 + V t, whereV = P /M . If there is an external force on the system so P = F ext, Then R satisfiesNewton’s second law MR = F ext.

For a two particle system the point R lies on a straight line joining the particles and isclosest to the heavier of the two particles. For a continuous body the center of mass is avolume integral

R =1

M

∫

dV rρ(r) (94)

where ρ is the mass density M =∫

dV ρ. Notice that if a large system is broken into twosubsystems we have

R =1

M

∑

k∈1

mkrk +1

M

∑

k∈2

mkrk =1

M(M1R1 + M2R2) (95)

where M1 =∑

k∈1 mk, M2 =∑

k∈2 mk are the total masses of the subsystems. For examplethe center of mass of a uniform sphere of radius a and mass m1 with a point mass m2 gluedsomewhere on its surface is the point on the line joining the point mass to the center of thesphere at radius m2R/(m1 + m2).


4.4 Angular Momentum

For a single particle angular momentum about the point with position vector R is definedby

L = (r − R) × p → r × p (96)

when it is about the origin of coordinates. We can use the 2nd law to find its time derivative

dL

dt= r × p + (r − R) × F = (r − R) × F ≡ N (97)

where N is called the torque about R. The first term vanished because p is parallel to r.We see that angular momentum about R is conserved if the torque about R vanishes. Forsimplicity we assume that R = 0, namely that we compute angular momentum and torqueabout the origin of coordinates.

An important situation in which the torque vanishes is if the vector F (r) is parallel tor. Such a force is called a central force. Examples include the gravitational and Coulombforces exerted by a mass or charge fixed to the origin of coordinates.

Conservation of angular momentum provides an elegant explanation of Kepler’s 2nd lawof planetary motion, which states that a single planet revolving about the sun sweeps outequal areas in equal times. first notice that if L is a constant vector, both r and p, whichat all times are perpendicular to the constant direction of L, lie in a fixed plane, and so theentire trajectory stays in that plane, which we take to be the xy-plane.

The area dA swept out in an infinitesimal time dt is simply the area of the triangledetermined by r and vdt. From our work with vectors, we have found that the area of aparallelogram spanned by any two vectors is |v1 × v2|, so the area of the triangle is just halfof this:

dA =1

2|r × v|dt =

|L|2m

dt (98)

Hence A = |L|/2m. Clearly, if L is conserved A is a constant, and Kepler’s second lawfollows. Notice that Kepler’s law is true for any central force, not just inverse square ones.In polar coordinates, we found that r = rr + rϕϕ so that

L = mr2ϕz (99)

When L is conserved, the angular velocity of the planet varies as 1/r2.

4.5 Systems of several particles

Let us consider a system of N particles and define the angular momentum about the originof coordinates as simply the sum of the angular momenta of the individual particles:

L =N∑

k=1

rk × pk (100)

dL

dt=

N∑

k=1

rk × F k = N total (101)


When we considered the time derivative of the total momentum, the 3rd law guaranteed thatthe internal forces canceled out so we had P = F ext, so that total momentum is conservedin the absence of external forces. The internal force contributions to the total torque do notautomatically cancel out, but they do simplify:

∑

k

rk ×∑

l 6=k

F kl =∑

k<l

rk × F kl +∑

k>l

rk × F kl

=∑

k<l

(rk × F kl + rl × F lk) =∑

k<l

(rk − rl) × F kl (102)

If all of the internal forces are central forces, which means F kl is parallel to (rk − rl) foreach pair k, l then the internal force contributions to the total torque do cancel.

dL

dt=

∑

k

rk × F k ext = N ext, Central Internal Forces (103)

Then in the absence of external forces the total angular momentum of the N particle systemwill be conserved. [Discuss Putty-turntable example]

The angular momentum of a system of particles about its center of mass has specialinterest. We might call this the “spin” of the system: think of a spinning top. Whenexternal forces are present, the center of mass coordinate R(t) will be accelerating accordingto MR = F ext, so the spin is not simply the angular momentum in another inertial system.With rk the coordinates relative to the origin of an inertial system, we define the coordinatesrelative to the center of mass as r′

k ≡ rk − R(t), and the momenta relative to the center ofmass are p′

k = mkr′k. Now write out L in terms of the primed variables:

L =∑

k

(r′k + R(t)) × (p′

k + mkR(t)

=∑

k

r′k × p′

k + R(t) ×∑

k

p′k +

∑

k

mkR(t) × R(t) +∑

k

mkr′k × R

=∑

k

r′k × p′

k + MR(t) × R ≡ S + R × P (104)

where M =∑

k mk is the total mass, and∑

k mkr′k =

∑

k mkrk −MR = 0 by the definition

of center of mass. Then also∑

k p′k = P − MR = 0. This result shows that we can

always decompose the total angular momentum of a system into its “spin” plus the angularmomentum of the system as a whole R × P .

If all the internal forces are central and obey the 3rd law, we then have

N ext =dL

dt=

dS

dt+ R × F ext

dS

dt= N ext − R × F ext =

∑

k

(rk − R) × F k ext (105)

In other words the angular momentum of a system of particles about its center of mass hasits time derivative equal to the torque about the center of mass, even if the center of mass

is accelerating. [Discuss dumbbell motion after sharp impulse]


5 Energy

The last great conservation law is that of energy. In mechanics we start with a precisenotion of Kinetic energy (KE for short). the kinetic energy of a particle of mass m is simplyT = mv2/2 = p2/2m. We can use Newton’s 2nd law to calculate its time derivative

dT

dt= v · F (106)

In general F can depend on position r, velocity r and time t. When it depends only onposition, we can formulate the concept of work and the work energy theorem. Integrate bothsides of the last equation w.r.t. time:

∆T = T (t) − T (0) =

∫ t

0

dt′dr

dt′· F (r(t′) =

∫

r(t)

r(0)

dr′ · F (r′) (107)

The integral on the right is an example of a line integral and can be defined for any pathconnecting the two points r1 = r(0) and r2 = r(t). To see this describe the chosen pathparametrically r(λ) = (x(λ, y(λ), z(λ), with r(0) = r1 and r(1) = r2. Then dr/dλ is avector tangent to the curve, and the line integral is defined as the ordinary integral

W (r1, r2; P ) =

∫

r2

P,r1

dr′ · F (r′) ≡∫ 1

0

dλdr

dλ· F (r(λ)) (108)

(In mechanics we can think of the parameter λ as time and the path as some trajectoryof the particle.) It is called the work W (r1, r2; P ) done by the force on the particle as itmoves from r1 to r2 along the path P . The previous equation shows that this work givesthe change in kinetic energy of the particle as it follows this motion.

The dependence of the work on the path used to define it limits its utility. Randomchoices for F are very likely to give path dependent work functions. Suppose for exampleF = (ay, bx, 0), and we calculate the work from (0, 0, 0) to (R,R, 0) in two ways.

W1 =

∫ R

0

dxFx(x, 0, 0) +

∫ R

0

dyFy(R, y, 0) = 0 + bR2 = bR2

W2 =

∫ R

0

dyFy(0, y, 0) +

∫ R

0

dxFx(x,R, 0) = 0 + aR2 = aR2 6= W1 (109)

Of course for the special case b = a the two paths give the same answer!

5.1 Conservative forces and potential energy

However, for the important class of forces, called conservative forces the work turns out tobe independent of the path. In that case, fixing r0 as some reference point, we can define afunction of all of space r by U(r) ≡ −W (r0, r). The reference point is the point at which


U(r0) = 0. If W depended on the path, there would be no way to consistently define sucha single valued function. When it can be defined this function is the potential energy.

When the force is conservative, the work done from r1 to r2 can be written

W (r1, r2) =

∫

r2

r1

dr · F =

∫

r0

r1

dr · F +

∫

r2

r0

dr · F

= U(r1) − U(r2) = −∆U (110)

Combining this with the work energy theorem then gives energy conservation ∆T +∆U = 0.Now let’s consider more closely the relation between U(r) and F (r). By definition

U(r) = U(x, y, z) = −∫

r

r1

(dxFx + dyFy + dzFz)

∂U

∂x= −Fx,

∂U

∂y= −Fy,

∂U

∂z= −Fz (111)

That is, the components of the vector F are the corresponding derivatives of U : Fi =−∂U/∂xi. Since there are three coordinates there are three derivatives that can be taken,each holding the other 2 constant. This is what is meant by the partial derivatives ∂/∂xi.These partial derivatives can be denoted by the vector gradient ∇:

∇ = x∂

∂x+ y

∂

∂y+ z

∂

∂z(112)

F = −∇U (113)

Forces which can be derived from a potential energy this way are precisely the conservativeforces.

It is a fundamental property of partial derivatives that the order in which two partialsare applied doesn’t matter:

∂2f

∂x∂y≡ ∂

∂x

(

∂f

∂y

)

=∂

∂y

(

∂f

∂x

)

=∂2f

∂y∂x(114)

and this is true for each pair, x, y, y, z, z, x. This means that

∂Fi

∂xj

= − ∂2U

∂xi∂xj

= − ∂2U

∂xj∂xi

=∂Fj

∂xi

∇iFj −∇jFi = 0 (115)

We can identify the combinations ∇iFj−∇jFi as the various components of the cross product∇× F , which is called the curl of F . Thus conservative forces have zero curl: ∇× F = 0.

5.2 Stoke’s Theorem

Another way to say that the work function is independent of the path is to say that the workdone along any closed path is zero:

∮

P

dr · F (r) = 0, All closed paths P (116)


We can relate the line integral on the right to a surface integral of the curl of F . To do thislet’s parameterize a surface by the three function rk(λ1, λ2). Where the parameters λ1, λ2

range over the unit square 0 ≤ λ1, λ2 ≤ 1. Then consider the integral∫ 1

0

dλ1dλ2

(

∂ri

∂λ1

∂rj

∂λ2

− ∂ri

∂λ2

∂rj

∂λ1

)

∂F i

∂rj

=

∫ 1

0

dλ1dλ2

(

∂ri

∂λ1

∂F i

∂λ2

− ∂ri

∂λ2

∂F i

∂λ1

)

=

∫ 1

0

dλ1dλ2

(

∂

∂λ2

[

∂r

∂λ1

· F]

− ∂

∂λ1

[

∂r

∂λ2

· F])

=

∫ 1

0

dλ1

[

∂r

∂λ1

· F] ∣

∣

∣

∣

λ2=1

λ2=0

−∫ 1

0

dλ2

[

∂r

∂λ2

· F] ∣

∣

∣

∣

λ1=1

λ1=0

= −∮

P

dr · F (117)

where P is the closed loop described by r(λ1, 0), 0 < λ1 < 1; r(1, λ2), 0 < λ1 < 1; r(λ1, 1),1 > λ1 > 0; r(0, λ2), 1 > λ2 > 0. Clearly this closed loop is the boundary of the surfaceparameterized by r(λ1, λ2). this is known as Stoke’s theorem. Our argument actually appliesin any number of spatial dimensions. In 3 dimensions we can write(

∂ri

∂λ1

∂rj

∂λ2

− ∂ri

∂λ2

∂rj

∂λ1

)

∂F i

∂rj

= ǫijkǫklm∇jFi ∂rl

∂λ1

∂rm

∂λ2

= −(∇× F ) ·(

∂r

∂λ1

× ∂r

∂λ2

)k

(118)

The infinitesimal vector dA = ∂r

∂λ1× ∂r

∂λ2dλ1dλ2 has magnitude equal to the element of area of

the parallelogram spanned by the two vectors ∂r

∂λ1dλ1 and ∂r

∂λ2dλ2, and direction perpendicular

to the surface with sign given by the right hand rule. So in 3 dimensions Stoke’s theoremcan be expressed as

∫

S

dA · ∇ × F =

∮

∂S

dr · F (119)

where ∂S is a standard notation for the boundary of the surface S. The sign is consistentwith the right hand rule: if the fingers of the right hand are pointed in the sense of thecontour on the right, the thumb points in the direction of dA.

With the aid of Stoke’s theorem, we see immediately that the work integral is independentof the path if and only if ∇ × F = 0 throughout all of space. Equivalently any force withzero curl can be expressed as minus the gradient of a potential.

5.3 Potential energy examples

• Central forces: U(r): F = −∇U = −U ′(r)r

r≡ f(r)r. Checking the curl:

∇× (f(r)r) =f ′(r)

rr × r + f(r)∇× r = 0 (120)

We see from the above the relation between U and F for central forces. We can alsodo the work integral explicitly

U(r) = −∫

r

r0

dr′ · r′f(r′) = −1

2

∫

r

r0

d(r′)2f(r′) = −∫ r

r0

r′f(r′)dr′ (121)

from which we see that U depends only on r = |r| and that U ′ = −rf(r).


• Uniform force (F = constant). U = −r · F .

−∇i(−rjF j) = δijFj = F i (122)

U(r) = −∫

r

r0

dr′ · F = −F · (r − r0) (123)

• Several distinct forces. Each conservative force can be expressed F k = −∇Uk. then thetotal potential energy U =

∑

k Uk makes sense. This can’t be done for nonconservativeforces F nc. Then energy conservation is replaced by ∆(T + U) = Wnc. Example:Inclined plane with friction.

5.4 One dimensional systems

With one dimensional systems, one only needs to require that the force is a function only ofx, F (x), not of t nor of v. Then the path independence of the work function is automatic,and one simply defines the potential energy by

U(x) = −∫ x

x0

dx′F (x′), F (x) = −dU

dx(124)

The force is zero whenever the curve U(x) has zero slope, i.e. at relative maxima or minima.At those points the particle can be in static equilibrium. At a maximum the equilibrium isunstable. A particle in motion has energy larger than U(x).

Energy conservation allows us to express the solution of a general one dimensional conser-vative force problem in terms of an explicit integral. We start with the statement of energyconservation:

1

2mv2 + U(x) = E (125)

with E the total (kinetic + potential) energy, which is a constant. We then solve for v as afunction of x:

v =dx

dt= ±

√

2(E − U(x))

m, ±

∫ t

0

dt′ =

√

m

2

∫ x

x0

dx′

√

E − U(x′)(126)

The sign is fixed by initial conditions and will change at turning points. Let’s choose the +sign from now on:

t =

√

m

2

∫ x(t)

x0

dx′

√

E − U(x′)(127)

Here we get t(x) as an explicit function of x, rather than x as an explicit function of t. Butit nevertheless tells us everything about the motion! To go further, we have to specialize tospecial potentials.


Let’s do the integral for a uniform force, such as gravity near the earth’s surface. ThenF = −mg or U(z) = mgz and

t = ±√

m

2

∫ z(t)

z0

dz′√E − mgz′

= ∓ 2

mg

√

m

2

[

√

E − mgz(t) −√

E − mgz0

]

(128)

If the particle starts out at rest, E = 0+mgz0 = mgz0, so the second term is zero we shouldchoose the − sign because the subsequent velocity will be negative. Then

t =

√

2

g

√

z0 − z(t), z(t) = z0 −1

2gt2 (129)

a result we are of course very familiar with.Systems can be effectively one-dimensional without literally involving motion along a

straight line. This can happen if three dimensional motion is suitably constrained–thinkof a roller coaster. Or consider a simple pendulum consisting of a massless rod of length Lconnected to a bob mass m. The motion is constrained to a circle of radius L. In the absenceof friction energy is conserved. The position of the bob is given in terms of the angle θ therod makes with the vertical: v = Lθ and z = L(1 − cos θ)

E =m

2v2 + mgz =

mL2

2θ2 + mgL(1 − cos θ) =

mL2

2θ2 + 2mgL sin2 θ

2(130)

Note that we can quickly recover the e.o.m. by setting E = 0: mL2θ + mgL sin θ = 0. Thewe apply the 1-d formula

t =

√

mL2

2

∫ θ

θ0

dθ′√

E − mgL(1 − cos θ′)(131)

Taking θ0 = 0 and noting that the maximum angle occurs when θ = 0, E = 2mgL sin2 θm

this simplifies to

t =

√

L

4g

∫ θ

0

dθ′√

sin2(θm/2) − sin2(θ′/2)(132)

the period T of the pendulum is 4 times the time it takes θ to rise from 0 to θm:

T = 2

√

L

g

∫ θm

0

dθ′√

sin2(θm/2) − sin2(θ′/2)= 4

√

L

g

∫ 1

0

du√1 − u2

√1 − k2u2

(133)

where k = sin(θm/2).Other examples: (1) Cube balanced on a cylinder; (2) Atwood machine. Constraints

make the motion effectively one dimensional. The key ingredient is that the constraints dono work: normal constraining forces are perpendicular to the motion, and friction is eitherabsent or does no work as with rolling without slipping.


5.5 Central forces

Central conservative forces are derived from a potential U(r) which depends only on thedistance from the center of force, which we choose to be the origin of coordinates. Theconserved energy is then

E =m

2v2 + U(r) (134)

With motion in three dimensions energy conservation is less powerful than in one dimension,because the direction of v is not constrained by it. It is convenient to use spherical polarcoordinates to describe the motion in central potentials. These are

r = r(sin θ cos ϕ, sin θ sin ϕ, cos θ) (135)

Then the velocity is fairly complicated:

v = rr + r sin θϕ(− sin ϕ, cos ϕ, 0) + rθ(cos θ cos ϕ, cos θ sin ϕ,− sin θ)

= rr + r sin θϕϕ + rθθ (136)

v2 = r2 + r2 sin2 θϕ2 + r2θ2 (137)

Here we have introduced the mutually orthogonal unit vectors r, θ, and ϕ. It is easy tocheck that r × θ = ϕ, r × ϕ = −θ,and θ × ϕ = r.

Incidentally, the gradient operator in spherical coordinates is complicated by the factthat the unit vectors are not constant: We can infer from

dr · ∇ = dr∂

∂r+ dθ

∂

∂θ+ dϕ

∂

∂ϕ(138)

and dr = drr + r sin θdϕϕ + rdθθ that

∇ = r∂

∂r+ ϕ

1

r sin θ

∂

∂ϕ+ θ

1

r

∂

∂θ(139)

which makes it obvious that −∇U can be a central force only if U is independent of angles!Returning to energy conservation we have in terms of spherical coordinates

E =m

2(r2 + r2 sin2 θϕ2 + r2θ2) + U(r) (140)

The angular velocities are not constrained by energy conservation, but remember that wealso have angular momentum conservation!

L = mr × v = mr × (r sin θϕϕ + rθθ) = mr2(− sin θϕθ + θϕ) (141)

L2 = m2r4(sin2 θϕ2 + θ2) (142)


We notice that the angular velocities in E occur exactly in the same combination as they doin L2 ≡ L2. Since L2 is a constant, we can safely eliminate the angular velocities in E:

E =m

2r2 +

L2

2mr2+ U(r) ≡ m

2r2 + UL(r) (143)

UL(r) = U(r) +L2

2mr2(144)

The radial coordinate can now be treated as in one dimensional motion. In particular wecan solve for t as a function of r as an explicit integral. By studying the graph of UL(r) wecan also map out all of the qualitative motions.

Since the direction of L is a constant, it is convenient to choose our polar axis (z-axis)parallel to L, which means that θ = π/2 and L = mr2ϕz. The particle moves in the xyplane (θ = π/2) and its angular velocity is related to L and r:

ϕ =L

mr2(145)

r =

√

2(E − UL(r))

m,

dϕ

dr=

L

r2

√

1

2m(E − UL(r))(146)

We can integrate the last equation to get ϕ(r):

ϕ = L

∫ r

r0

dr′

r′2

√

1

2m(E − UL(r′))= L

∫ 1/r0

1/r

du

√

1

2m(E − UL(1/u))(147)

For the −C/r potential UL(1/u) = L2u2/2m − Cu and the integral is an elementary oneinvolving an inverse trig function. The turning points are given by

0 = L2u2/2m − Cu − E, u± =C ±

√

C2 + 2EL2/m

L2/m≡ 1 ± e

p(148)

p =L2

mC, e =

√

1 +2EL2

mC2(149)

For an actual turning point r = 1/u must be positive. We see that u± are both positivewhen C > 0 (attraction) and E < 0. In this case r oscillates between a minimum rp = 1/u+

(perihelion) and a maximum (aphelion) ra = 1/u−. When C > 0 and E > 0, only u+ ispositive, and the particle motion is unbounded. When C < 0 (repulsion), then necessarilyE > 0 and again only u+ is positive and the motion is unbounded. For gravity C = Gm1m2 >0. For the Coulomb potential C = −q1q2/4πǫ0 is negative for like sign charges and positivefor opposite sign charges.


To do the integral, we write

E + Cu − L2u2

2m= − L2

2m(u − u+)(u − u−) =

L2

2m

(

e2

p2−[

u − 1

p

]2)

(150)

ϕ =

∫ u+

1/r

du

√

1

e2/p2 − [u − 1/p]2

= arcsin

(

pu+

e− 1

e

)

− arcsin

(

p

re− 1

e

)

=π

2− arcsin

(

p

re− 1

e

)

cos ϕ =p

re− 1

e, r =

p

1 + e cos ϕ(151)

When the eccentricity e < 1 (so E < 0), the radial coordinate r varies between a minimump/(1 + e) and a maximum p/(1 − e). The orbit is then an ellipse with semi-major axisa = p/(1−e2). Clearly when e = 0 the radius is constant so the orbit is a circle. When e > 1(so E > 0) there is a minimum p/(1 + e) but no maximum: r → ∞ when cos ϕ → −1/e.

To find the time dependence of the motion we need

r =

√

2

m

√

E +C

r− L2

2mr2=

1

mr

√2mEr2 + 2mCr − L2

=

√−2mE

mr

√

(rmax − r)(r − rmin) (152)

When the orbit is bounded (E < 0), it has a period given by

T = 2

∫ rmax

rmin

dr

r= 2

m√−2mE

∫ rmax

rmin

rdr√

(rmax − r)(r − rmin)

= 2m√

−2mE

∫ 1

0

(rmax − rmin)xdx + rmindx√

x(1 − x)

=m(rmax + rmin)√

−2mE

∫ 1

0

dx√

x(1 − x)=

mπ(rmax + rmin)√−2mE

(153)

Now

2a ≡ rmax + rmin = 2p/(1 − e2) = 2L2

mC

mC2

(−2EL2)=

C

−E

T =2πa3/2

√

C/m,

T 2

a3=

4π2m

C(154)

which is Kepler’s third law.


5.6 Energy of systems of particles

Now we would like to extend the concepts of kinetic energy and potential energy to morethan one particle. The total kinetic energy of several particles is simply the sum of thekinetic energies of each particle:

T =N∑

k=1

1

2mkr

2k (155)

The definition of potential energy is less obvious. If the particles do not interact with eachother but each particle k moves independently in an external force field F k(r) then the forceon particle k is just F k(rk). If the force is conservative F k(r) = −∇Uk(r) and the force onparticle k is

F (rk) = −∇kUk(rk) = −∇k

N∑

l=1

Ul(rl) ≡ −∇kUT , UT =N∑

l=1

Ul(rl) (156)

where we understand that ∇k involves only derivatives w.r.t. the components of rk, holdingconstant the components of all rl with l 6= k. Thus all terms in the sum with l 6= k contributenothing to the force on particle k. With these definitions of total kinetic and potential energy,we have energy conservation:

E =N∑

k=1

(

1

2mkr

2k + Uk(rk)

)

(157)

dE

dt=

N∑

k=1

(mkrk · rk + rk · ∇kUk(rk)) =N∑

k=1

rk · (mkrk + ∇kUk(rk))

=N∑

k=1

rk · (mkrk − F k(rk)) = 0 (158)

by Newton’s second law for each particle.Next we want to include interactions between particles. Consider first two particles with

position vectors r1, r2. By Newton’s third law the mutual forces are equal and oppositeF 12 = −F 21. We assume that the mutual forces are translationally invariant. This meansthat they are the same when the particles are at positions r1 + a, r2 + a as they are atpositions r1, r2. This means that the mutual force depends only on the difference r1 − r2.

F 12 = F 12(r1 − r2) = −F 21 = −F 12(r2 − r1) (159)

The mutual gravitational and Coulomb forces are important examples. For example New-ton’s law of mutual gravity can be written

Fgrav12 = −Gm1m2

r1 − r2

|r1 − r2|3= −F 21 (160)


The crucial feature here is that the third law is included in the statement by the oddness ofF grav as a function of its argument.

If the mutual two-body force is conservative F 12(r) = −∇U12(r) where r = r1−r2. Theoddness of F 12 implies that U12(r) = U12(−r). Interestingly, we can write F 12 in three ways

F 12 = −∇U12 = −∇1U12 = +∇2U12 = −F 21 (161)

Thus we have the intuitive results F 12 = −∇1U12 and F 21 = −∇2U12. It is important toappreciate that the single potential energy U12 accounts for both the force of particle 2 onparticle 1 and the force of particle 1 on particle 2. For this reason the total potential energydescribing external forces F k and the mutual forces F 12 = −F 21 is written

UT = U1(r1) + U2(r2) + U12(r1 − r2) (162)

dE

dt= r1 · (m1r1 + ∇1U1) + r2 · (m2r2 + ∇2U2) + (r1 − r2) · ∇1U12

= r1 · (m1r1 + ∇1U1 + ∇1U12) + r2 · (m2r2 + ∇2U2 + ∇2U12)

= r1 · (m1r1 − F 1 − F 12) + r2 · (m2r2 − F 2 − F 21) = 0 (163)

by Newton’s second law applied to each particle.The iconic application of two body energy conservation is elastic collisions. It is based

on the fact that physical potential energies vanish at large separation. Thus the total energyE = T + UT → T when r1, r2, r1 − r2 all become large. In a scattering process, the twoparticles are initially very far from each other and from the center of external forces, soinitially E ≈ Ti. After scattering the two particles are again very far from each other andfrom the center of external forces, so E ≈ Tf . But E is constant through out the motion,so we have Tf = Ti for an elastic two body scattering process. If the external forces are allzero, then Newton’s third law implies that total momentum is also conserved. In this lattercase the initial and final variables are subject to the constraints

p1 + p2 = p′1 + p′

2 (164)1

m1

p21 +

1

m2

p22 =

1

m1

p′21 +

1

m2

p′22 (165)

The most efficient way to impose these constraints is to work in the center of mass systemfor which p1 +p2 = 0. Then momentum conservation simple states that p′

2 = −p′1. Plugging

these relations into the energy conservation equation then shows that p′21 = p2

1. That is p1

and p′1 have the same length. Thus p′

1 ·p1 = p21 cos θ, and the scattering process is completely

determined by the scattering angle θ. If the two particles have the same mass m1 = m2,these statements apply to the velocities and speeds of all the particles. Then the lab frame,in which v2 = 0, moves with velocity −v1 relative to the center of mass. Then the final stateparticles have velocities v′

1 + v1 and −v′1 + v1 and their scalar product is

(v′1 + v1) · (−v′

1 + v1) = v12 − v1

′2 = 0 (166)

So in the equal mass case the final state particles make tracks perpendicular to each other.


Finally, we want to consider a general N particle system. We shall assume that themutual forces are all two body forces. This means that the force on each particle k is a sumof forces from each of the other particles plus an external force

F k = F k(rk) +∑

l 6=k

F kl(rk − rl) (167)

where we also assume that the mutual forces are translationally invariant. One could imaginea more complicated situation with multi-body forces, but this will not be necessary. In thisnotation Newton’s third law is the statement that F lk(rl − rk) = −F kl(rk − rl). Newton’slaw of universal gravity

Fgravkl (rk − rl) = −Gmkml

rk − rl

|rk − rl|3(168)

clearly satisfies these requirements. The meaning of conservative forces immediately extendsfrom our two particle considerations. Namely,

F kl = −∇kUkl(rk − rl) = +∇lUkl(rk − rl) (169)

where the third law requires that Ulk(rl − rk) = Ukl(rk − rl). Just as in the two particlecase, the total potential energy will contain only one contribution for each pair:

Utot =∑

k

Uk(rk) +∑

k<l

Ukl(rk − rl) (170)

The justification for this choice is that it is the one that guarantees energy conservation:

dE

dt=

∑

k

rk · [mkrk + ∇kUk(rk)] +∑

k<l

(rk − rl) · ∇kUkl(rk − rl)

=∑

k


k<l

rk · ∇kUkl(rk − rl) −∑

k<l

rl · ∇kUkl(rk − rl)

=∑

k


k<l

rk · ∇kUkl(rk − rl) +∑

k<l

rl · ∇lUkl(rk − rl)

=∑

k


k<l

rk · ∇kUkl(rk − rl) +∑

l<k

rk · ∇kUkl(rk − rl)

=∑

k


l 6=k

rk · ∇kUkl(rk − rl)

=∑

k

rk · [mkrk + ∇kUk(rk) + ∇k

∑

l 6=k

Ukl(rk − rl)]

=∑

k

rk · [mkrk − F k] = 0 (171)

by Newton’s second law for each particle.


It is important to appreciate that energy conservation will follow if the potential energy isan arbitrary function of the coordinates of the system, as long as it has no explicit dependenceon time. This is just a consequence of the chain rule:

dU

dt=

N∑

k=1

rk · ∇kU(r1, r2, · · · , rN) = −N∑

k=1

rk · F k (172)

dE

dt=

∑

k

rk · (mkrk − F k) = 0 (173)

by Newton’s second law. Furthermore, we will also have momentum conservation if

0 =∑

k

F k = −∑

k

∇kU(r1, r2, · · · , rN) (174)

which is true if U(r1 + a, r2 + a, · · · , rN + a) = U(r1, r2, · · · , rN). In other words, momen-tum conservation holds if the potential energy is invariant under a uniform translation ofcoordinates.

6 Oscillations

Our next subject is to study the physics of harmonic oscillations. Recall Hooke’s law for aone dimensional oscillator F (x) = −kx. This force can be derived from the potential energyU(x) = kx2/2. We will spend a lot of time studying the physics of this system, but first it isimportant to appreciate the universality of Hooke’s law. It will come into play whenever onehas a point of stable equilibrium. Equilibrium requires that the force is zero or U ′(x0) = 0.Stability requires that U ′′(x0) > 0. whatever function U(x) is, we can always do a Taylorexpansion about a point x0:

U(x) = U(x0) + (x − x0)U′(x0) +

1

2(x − x0)

2U ′′(x0) + O(x − x0)3

→ U(x0) +k

2(x − x0)

2 + O(x − x0)3 (175)

for an equilibrium point and where k ≡ U ′′(x0) > 0 for stability. As long as the motion stayssufficiently close to x0, the dynamics can be well approximated by Hooke’s law! A simpleexample that we have already discussed is the motion of a pendulum in small oscillations:

U(θ) = mgL(1 − cos θ) ≈ mgL

2θ2 (176)

For small oscillations the energy is then

E ≈ 1

2ml2θ2 +

mgL

2θ2 (177)

From which energy conservation gives Hooke’s law θ = −(g/L)θ.


6.1 Descriptions of Simple Harmonic Motion

We can put the equation of motion for harmonic oscillations in the generic form

x = −ω2x (178)

where the angular frequency ω depends on the system: e.g.√

k/m for a spring constant k,√

g/L for a pendulum.We already know the general solution is a linear combination of trig functions cos ωt and

sin ωt, but there are several convenient ways to “package the solutions. One popular way isusing complex exponentials e±iωt:

x(t) = Ceiωt + C∗e−iωt (179)

where C is any complex number and C∗ is the complex conjugate of C, so that x(t) is a realnumber. We can easily relate C to the initial conditions:

x(0) = C + C∗, x(0) = iω(C − C∗), C =1

2

(

x(0) − ix(0)

ω

)

(180)

We can of course use Euler’s formula e±iωt = cos ωt± i sin ωt to rewrite the solution in termsof trig functions

x(t) = x(0) cos ωt +x(0)

ωsin ωt (181)

However we write the solution, it is clear that the motion is periodic with period T = 2π/ω.There is one more useful way to describe the general motion in terms of amplitude and

phase:

x(t) = A cos(ωt − δ), x(0) = A cos δ, x(0) = Aω sin δ

A =√

x(0)2 + x(0)2/ω2, tan δ =x(0)

ωx(0)(182)

This last representation has a neat connection with the complex exponential representation:put C = (A/2)eiδ. Then

x(t) =A

2

(

ei(δ−ωt + e−i(δ−ωt)

= ReAeiδe−iωt = A cos(ωt − δ). (183)

If we understand that we will always take the real part of any complex solution at the end,there is no harm in working with complex solutions from the beginning. This will be anextremely useful point of view when we consider damping.

Finally it is useful to relate the various parameterizations of simple harmonic motion tothe conserved energy of the system.

E =1

2mx(t)2 +

1

2kx(t)2 =

1

2mω2A2 sin2(ωt − δ) +

1

2kA2 cos2(ωt − δ) =

1

2kA2 (184)

So the energy is proportional to the square of the amplitude of motion. The total energy isconserved as it oscillates between kinetic and potential energy with angular frequency 2ω.


6.2 2 and 3 dimensional oscillations

We can look for stable equilibrium points in higher dimensional potentials U(r). Equilibriumrequires that ∇U = 0. The multidimensional Taylor expansion reads

U(r) = U(r0) + (r − r0) · ∇U(r0) +1

2(ri − ri

0)(rj − rj

0)∇i∇jU(r0) + · · ·

→ U(r0) +1

2kij(r

i − ri0)(r

j − rj0) + · · · (185)

where kij = ∇i∇jU(r0) is a 3× 3 spring constant matrix. For now we will assume that thismatrix is diagonal with positive diagonal elements when the equilibrium is stable:

k =

kx 0 00 ky 00 0 kz

(186)

Taking the equilibrium point as our origin of coordinates and the zero of potential energy tobe the equilibrium value, we then have

U(x, y, z) =1

2

(

kxx2 + kyy

2 + kzz2)

+ · · · (187)

We see that the total energy will then be a sum of three independently conserved terms

E =1

2(mx2 + kxx

2) +1

2(my2 + kyy

2) +1

2(mz2 + kzz

2) (188)

In principle the angular frequency of each coordinate will be different ωx,y,z =√

kx,y,z/m.If all spring constants are equal we have the isotropic oscillator, with potential energy

U = (k/2)(x2 + y2 + z2) = kr2/2. In that case the force will be central F = −kr andangular momentum will be conserved. As we know this implies that the motion is in a planeperpendicular to L, and we can assume this plane is the xy-plane. then the general solutioncan be written

x(t) = Ax cos ωt, y(t) = Ay cos(ωt − δ) = Ay cos δ cos ωt + Ay sin δ sin ωt (189)

(y(t) − x(t)Ay cos δ/Ax)2 = A2

y sin2 δ sin2 ωt (190)

1 =x2

A2x

+(y(t) − x(t)Ay cos δ/Ax)

2

A2y

(191)

which is the equation for a rotated ellipse. When δ = π/2, the equation reduces to thestandard form x2/a2 + y2/b2 = 1 with a = Ax and b = Ay. If δ = 0, x = (Ax/Ay)y and theelliptical orbit. collapses to a straight line.

The motion gets quite interesting when the spring constants in different directions aredifferent. If ωx/ωy is rational the resulting motion will eventually be periodic although xand y may have to go through many periods for the motion to repeat. For example ifωx = 2ωy, x executes 2 periods as y executes only one. However if the ratio of frequencies isirrational the two dimensional motion will never repeat: this non-repeating motion is termedquasi-periodic.


6.3 Damped Oscillations

In the real world no system will oscillate forever because of inevitable frictional forces.Fortunately it is easy to incorporate damping and keep the equations of motion linear. Wesimply take the damping force proportional to the velocity F damp = −bx ≡ −2mβx, withβ > 0 so that it impedes motion. The equation of motion for a single damped oscillator isthen

x + 2βx + ω20x = 0. (192)

This e.o.m. is a special case of a linear differential equation with constant coefficients.Let’s take a brief mathematical excursion to consider how any such equation can be

handled. The key ingredient is to find a complete set of solutions, each of which diagonalizesthe derivative operator d/dt. We are of course very familiar with the functions that do this:The exponential functions ert. Let us denote a general differential operator by the symbolD:

Dx =

(

a0 + a1d

dt+ a2

d2

dt2+ · · · + an

dn

dtn

)

x =n∑

k=0

akdk

dtk(193)

Then

Dert =n∑

k=0

akrkert = P(r)ert (194)

In other words the differential equation Dx = 0 is solved by x(t) = Cert provided r isa root of the polynomial P . We have converted a problem in differential equations to aproblem in algebra! To get the complete solution to the differential equation we need nindependent solutions. Fortunately the fundamental theorem of algebra guarantees thatthere are precisely n complex roots of any nth order polynomial. It is of course crucial thatwe allow complex roots for this theorem to hold. It is also important that some of the nroots can be repeated. For example the two roots of the equation (r − 1)2 = 0 are bothequal to 1. When tire are multiple roots, we can get several solutions by multiplying ert bya polynomial of t. For example if r is a double root, it means that D contains the factors

(

d

dt− r

)2

(195)

applying these operators to (a + bt)ert we find

(

d

dt− r

)2

(a + bt)ert =

(

d

dt− r

)

bert = 0 (196)

More generally

(

d

dt− r

)k

(b0 + b1t + · · · + bk−1tk−1)ert = 0 (197)


because the polynomial factor can prevent only k − 1 of the factors from giving zero.With this background we return to our damped oscillator e.o.m. for which the polynomial

is quadratic

r2 + 2rβ + ω20 = 0, r± = −β ±

√

β2 − ω20 (198)

and the general solution of the equation is

x(t) = e−βt(

C1et√

β2−ω20 + C2e

−t√

β2−ω20

)

(199)

The qualitative nature of these solutions depends dramatically on the relationship betweenβ and ω0. If β < ω0 (underdamping), we can write

√

β2 − ω20 = −i

√

ω20 − β2 ≡ −iω1 and

the solution reads

x(t) = e−βt(

C1e−itω1 + C∗

1eitω1)

(200)

and we see that we have oscillations with frequency ω1 < ω0 and amplitudes that decayexponentially in time. Note that in this case we choose C2 = C∗

1 to make the solution real.In contrast when β > ω0 (overdamping) the square root is real and both terms are

exponentially damped at different rates

x(t) = C1e−β1t + C2e

−β2t (201)

β1 = β −√

β2 − ω20, β2 = β +

√

β2 − ω20 (202)

with the first term dying off least slowly.For β = ω0 (critical damping), the two roots are the same (the polynomial has a double

root) so the two independent solutions are e−βt and te−βt. Notice that in overdampedand critically damped cases, damped oscillation is a misnomer since not even one cycleof oscillation is completed. There can be at most one time when the amplitude vanishes,and that is possible only when the two solutions occur with opposite signs in the linearsuperposition.

6.4 Driving oscillations

Given the reality of damping forces, in order to keep the system oscillating, a driving forcemust be applied. For example, a pendulum clock receives a “kick” with each tick, poweredby gravity (via weights) or perhaps by a spring, or perhaps by a battery. Of course the powersource must be regularly renewed.

Sinusoidal driving forces are the easiest to solve. We can put F = mf0 cos ωt = mRe f0e−iωt.

It is in fact most efficient to use the time dependence e−iωt, taking the real part at the endof the calculation. Then the e.o.m. to solve is simply

x + 2βx + ω20x ≡ Dx = f0e

−iωt (203)


This is what we call an inhomogeneous linear differential equation. This is because thedriving term does not involve x. The equation with f0 = 0 is called a homogeneous differentialequation. Suppose we have found some particular solution xp(t) to the inhomogeneousequation. This solution usually not obey the desired initial conditions. But we can add toxp any solution xh to the homogeneous equation Dxh = 0 and get another general solution:

D(xp + xh) = Dxp + Dxh = Dxp + 0 = f0e−iωt. (204)

With the two free parameters in xh we have the flexibility to impose any initial conditionson x(0) and x(0).

So the problem is to find a particular solution. Recall that the differential operator Dapplied to an exponential ert just multiplies the exponential by a polynomial in r:

DAert = P(r)Aert (205)

Thus, by choosing r = −iω, the time dependence of both sides of the equation matches andcan be cancelled out leaving the equation

P(−iω)A = f0, A =f0

P(−iω)(206)

x(t) = xp(t) + xh(t) = xh(t) +f0

P(−iω)e−iωt

= xh(t) + Ref0

ω20 − ω2 − 2iβω

e−iωt (207)

In order for this construction to work it is important that P(−iω) 6= 0. When there isnonzero damping β > 0, P is never zero for real ω. At the same time the solution xh isexponentially damped e−βt, so that after a sufficiently long time βt ≫ 1 the xh term diesaway and the system settles down to a steady state of motion

x(t) ∼ Ref0

ω20 − ω2 − 2iβω

e−iωt (208)

Before discussing this damped driven solution, let us briefly consider what happens in theabsence of damping β = 0. For ω 6= ω0, the solution is much the same except the xh term nolonger dies out with time. However when ω = ω0 in the undamped case this method breaksdown, and one has to go back to the original diff eq to get xp.

x + ω20x = f0e

−iω0t (209)

The problem is that xp = Ae−iω0t gives 0 on the left side. However, simply multiplying by tfixes the problem:

d2

dt2(te−iω0t) = −ω2

0te−iω0t − 2iω0e

−iω0t (210)


so we solve the inhomogeneous equation with xp = Ate−iω0t and A = if0/(2ω0). This is asimilar phenomenon to our discussion of critical damping. The xh term doesn’t die away,but the xp term grows relative to the xh term linearly in t. So it is still true that the xp termwill eventually dominate the solution.

We now return to the damped case where this subtlety is avoided. It is interesting thatafter the xh term (“transients”) dies away the solution is uniquely determined by the drivingterm. Let’s assume that f0 is real and positive. Then we can write

f0

ω20 − ω2 − 2iβω

=f0

√

(ω20 − ω2)2 + 4β2ω2

eiδ ≡ Aeiδ tan δ =2βω

ω20 − ω2

(211)

In summary the driving force F = mf0 cos ωt uniquely produces the steady state solution(after transients have died away)

xsteady = A cos(ωt − δ) (212)

We sometimes call xsteady the response to the driving force. Its time dependence is out ofphase with the driving force, lagging by the time δ/ω. Both A and δ depend sensitively onthe relation of ω to ω0. When ω = ω0 tan δ blows up implying that δ → π/2. For smalldamping A can get very large at that frequency.

6.5 Resonance

As we have seen, the driven damped harmonic oscillator is, for all practical purposes, de-scribed by its steady state motion after transients have died away. For sinusoidal drivingforce, the response is sinusoidal with the same frequency but with a phase shift. The strengthof the response, as measured by the squared amplitude (which is proportional to the powerdelivered to the system) is completely fixed:

A2(ω) =f 2

0

(ω20 − ω2)2 + 4β2ω2

(213)

The power delivered to the system is the rate at which the driving force does work:

P = xF = −ωAmf0 sin(ωt − δ) cos ωt = −ωAmf0(sin ωt cos ωt cos δ − cos2 ωt sin δ

〈P 〉 =1

2ωAmf0 sin δ =

1

2ωAmf0

2βω√

(ω20 − ω2)2 + 4β2ω2

=mβω2f 2

0

(ω20 − ω2)2 + 4β2ω2

(214)

The actual motion of the oscillator is less interesting that the the way the response dependson the frequency. In spection of the preceding formula shows that the power delivered to theoscillator is a maximum when ω = ω0. It is perhaps not surprising that the biggest responsecomes when driven at the natural frequency: we say that the system is in resonance. Theresonance phenomenon is most dramatic for weak damping, β ≪ ω0. Then the resonanceresponse gets very large:

A2(ω0) =f 2

0

4β2ω20

≫ f 20

ω40

(215)


The resonance peak stands very tall compared to the non resonant amplitude. (Whenω2 > 2ω2

0, A2 is smaller than f 20 /ω4

0.)With weak damping, the resonance peak is not only very high, but it is also very narrow.

A convenient measure of the width of the resonance peak is the value of ω − ω0 where theresponse is half of the maximum value. Taking the power delivered as a good measure ofthe response, this condition gives

|ω2 − ω20| = 2βω, ω = β +

√

ω20 − β2 ≈ ω0 + β (216)

So for weak damping the width of the peak at half maximum is 2β ≪ ω0. if the name of thegame is to have the sharpest possible peak (say to tune in a desired radio station) then theratio Q = ω0/(2β) (the Quality factor) should be as large as possible, Q ≫ 1. Notice thatQ is a parameter characteristic of the damped oscillator: the properties of the driving forcedon’t enter into it. Q is simply proportional to the ratio of the decay time of the unforcedoscillator to its period: high Q means that the oscillations will last many periods beforedying away. high Q is synonymous with a long lifetime.

So far we have studied the strength of the response at resonance. The other feature ofthe response is the phase shift δ determining the lag time δ/ω of the response compared tothe driving force. We have noted that tan δ blows up at ω = ω0 implying that δ → π/2at resonance. This means the response and driving force are 90◦ out of phase: when oneis a maximum the other is zero. When ω < ω0 and β is small δ is also small. It risesrapidly to π/2 at ω = ω0. For ω > ω0, tan δ becomes small and negative, implying that δ isapproximately π. So the behavior of the phase shift is a rapid rise through π/2 from near 0to π as ω increases past resonance.

6.6 Fourier Series

We have seen that sinusoidal driving forces are particularly easy to handle. Because theequations are linear we can solve any linear combination of sinusoidal driving forces bysimply taking the corresponding linear combination of solutions:

f(t) =∑

k

fke−iωkt, x =

∑

k

fk

ω20 − ω2

k − 2iβωk

e−iωkt (217)

For example the periodic kicks given to the pendulum of a clock are by no means sinusoidal.Let’s find the condition on the ωk for f to be periodic of period τ , f(t + τ) = f(t).

e−iωk(t+τ) = e−iωkt, e−iωkτ = 1, ωk =2nπ

τ≡ nω (218)

where n is any integer. We might as well use n as our summation index and so we write

f(t) =∞∑

n=−∞

ane−inωt, x =

∞∑

n=−∞

an

ω20 − n2ω2 − 2iβnω

e−inωt (219)


This equation displays the Fourier series representation of f(t). The powerful fact is thatsuch a Fourier series can described practically any periodic function of t. Even better wecan get a formula for each fn as an integral. This is based on the following property:

∫ τ

0

dtei(m−n)ωt =ei(m−n)ωτ − 1

i(m − n)ω= 0, for m 6= n (220)

Clearly when m = n this integral just gives τ . Then we find

∫ τ

0

dtf(t)eimωt =∞∑

n=−∞

an

∫ τ

0

ei(m−n)ωt = τam

am =1

τ

∫ τ

0

dtf(t)eimωt (221)

As an important example, we find the coefficients for a square pulse f(t) = f0 for 0 < t <ǫ < τ and f(t) = 0 for ǫ < t < τ . Then a0 = f0ǫ/τ , and for m 6= 0,

am =f0

τ

∫ ǫ

0

dteimωt =f0

imωτ(eimωǫ − 1) =

f0

2πim(eimωǫ − 1) =

f0

mπeimωǫ/2 sin

mπǫ

τ(222)

Notice that if we pretend m is continuous we have a0 = lim am as m → 0. Then

f(t) =ǫf0

τ+∑

n6=0

f0

nπsin

nπǫ

τe−inω(t−ǫ/2)

=ǫf0

τ+

∞∑

n=1

2f0

nπsin

nπǫ

τcos nω(t − ǫ/2) (223)

The particular solution describing the response to this force is simply the sum of responsesto each term:

xp(t) =ǫf0

τω20

+∑

n6=0

f0

nπsin

nπǫ

τ

1

ω20 − n2ω2 − 2iβnω

e−inω(t−ǫ/2)

=ǫf0

τω20

+ 2∞∑

n=1

f0

nπsin

nπǫ

τ

1√

(ω20 − n2ω2)2 + 4n2β2ω2

cos[nω(t − ǫ/2) − δn](224)

For weak damping β ≪ ω0, the nth denominator gets small when ω = ω0/n or τ = nτ0.when this happens we can say that the nth term is in resonance, and we can expect thatsingle term to be dominant.

6.7 RMS Displacement, Parseval’s Theorem

Sometimes it is good to have a single time independent measure of the response to a drivingforce. If we average x over a period we simply pick out the 0th term: 〈x〉 = ǫf0/(τω2

0). This


shows none of the resonance behavior. A better measure is the mean squared displacement:

〈x2〉 =∑

m,n

am

ω20 − m2ω2 − 2iβmω

an

ω20 − n2ω2 − 2iβnω

1

τ

∫ τ

0

dte−i(n+m)ωt

=∑

n

ana−n

(ω20 − n2ω2)2 + 4n2β2ω2

→∑

n

f 20

n2π2

sin2(nπǫ/τ)

(ω20 − n2ω2)2 + 4n2β2ω2

(225)

for the square pulse example. The RMS displacement is just the square root of this result,defined to give a measure with the dimensions of length instead of length squared. Thestrength of the nth resonance peak to the RMS Displacement is

√

〈x2〉n ∼ f0

2βω0nπsin

ǫω0

2(226)

decreasing like 1/n.

7 Variational Principles

Up to now we have formulated classical mechanics in terms of Newton’s three laws. Whenwe began to understand conservation laws, we learned that there were dramatic shortcuts tosolving the equations of motion. The special case of finding static solutions, we learned thatthey were simply the stationary points of the potential energy function U ′(x) = 0. Thoughtabout this way, it is clear that a minimum is a minimum, regardless of the coordinate choice.What we would now like to understand is how general solutions can be determined by asimilar principle, known as Hamilton’s principle of least action. The solutions are requiredto be stationary points of the action, defined for dynamics determined by a potential energyas

I =

∫ t2

t1

dt(T − V ) (227)

One can calculate I for any trajectory with initial and final conditions x(t1) = x1 andx(t2) = x2. But the actual solution is precisely the one which minimizes I. We need a newway of testing for minima w.r.t. small changes in the function x(t). This is the so-calledcalculus of variations.

7.1 Examples

Consider the question: what path gives the shortest distance between two points? To answerthis question we have to compare the distances given by different paths. We first need aformula for the distance on a fixed path P : To describe a path it is enough to give threefunctions r(λ) where λ marks a unique point on the path, say with r(0) = r1 and r(1) = r2.


This is just a parametric description of a curve in space. To calculate the distance we startwith the differential dr = dλr′(λ). Then ds = dλ

√r′2 and

D12 =

∫ 1

0

dλ√

r′2 (228)

The calculus of variations will then tell us how to minimize this distance.Another example of a variational principle is Fermat’s principle of optics: A light ray

follows a path that minimizes the time to complete the journey. The speed of light c/ndepends on the index of refraction n(r) which can vary in space. Then the time of thejourney is

T12 =

∫ 1

0

dλ|r′|c/n

=1

c

∫ 1

0

dλn(r(λ))√

r′2(λ) (229)

Hamilton’s Principle is exactly the same type of mathematical problem. We have to find afunction that minimizes a quantity that is expressed as the integral over some parameter ofa function of the path r and its derivative r′ over the extent of the path. For Hamilton’sprinciple the parameter is the time t, the path is the particle’s trajectory and the velocity isthe derivative.

So in complete generality we need to minimize an integral of the form

I =

∫ λ2

λ1

dλL(qk(λ), qk(λ), λ) (230)

Here we use a dot to denote derivative w.r.t. λ, which will be time for mechanics. Forpedagogical purposes we will just consider one coordinate in the following discussion. Thegeneralization to any number is easy.

7.2 Calculus of Variations

So we have to study how I changes under a small distortion of the path. Suppose we havea trial path qk(λ). Then we can write a nearby path as qk(λ) + δqk(λ). Here δq is small butarbitrary in shape. We can even choose δq to be zero over most of the path, departing fromthe path only in a tiny interval of λ:

∆I =

∫ λ2

λ1

dλ [L(qk(λ) + δqk(λ), qk(λ) + δqk(λ), λ) − L(qk(λ), qk(λ), λ)]

=

∫ λ2

λ1

dλ∑

l

[

δql∂L

∂ql

+ δql(λ)∂L

∂ql

]

+ O(δq2)

=∑

l

δql∂L

∂ql

∣

∣

∣

∣

λ2

λ1

+

∫ λ2

λ1

dλ∑

l

δql

[

∂L

∂ql

− d

dλ

∂L

∂ql

]

+ O(δq2) (231)


Now since the ends of the trajectory are fixed, δqk(λ1) = δqk(λ2) = 0, so qk(λ) will make Istationary if

d

dλ

∂L

∂ql

=∂L

∂ql

, for all l (232)

These are the Euler Lagrange equations Without saying so, we have just applied what isknown as the calculus of variations!

7.3 Nonmechanics applications of the calculus of variations.

Let us apply Lagrange’s equations to a couple of examples, starting with the geodesic problemin flat space. Recall that the distance between two points in 3 dimensional space is given by

D12 =

∫ 1

0

dλ√

r2 (233)

where now we use the dot to indicate derivative w.r.t. λ. So in this case L =√

r2 so weneed

∂L

∂rk

= 0,∂L

∂rk

=rk√r2

=rk

|r| (234)

The E-L equations just say that r

|r|, which is just the unit tangent vector to the path, is a

constant over the path. This means that the path is a straight line!Another famous example is the brachistochrone problem: What is the shape of the roller

coaster track between two points such that the frictionless car takes the minimum time tocomplete the journey? To answer this we need to know the speed of the car at any pointon the track. Take point 1, where the car starts at rest, to be the origin of Cartesiancoordinates and the gravitational potential energy to be U = mgz. Then by conservationof energy v =

√−2gz. Recall the element of arc length ds = dλ√

x2 + z2 so we need tominimize

T12 =

∫ λ2

λ1

dλ

√x2 + z2

√−2gz(235)

∂L

∂x= 0,

∂L

∂z= − L

2z,

∂L

∂x=

x√−2gz√

x2 + z2,

∂L

∂z=

z√−2gz√

x2 + z2(236)

We are free to choose λ in a way to simplify the equations. In the text the choice λ = z ismade. For variety, let’s try choosing λ so that

√x2 + z2 = 1. Then the E-L equations tell

us that

∂L

∂x=

x√−2gz= C, z = −

√

1 + 2C2gz (237)


where we chose the negative square root because initially z decreases with λ. Integratingthe second equation gives

λ = −∫

dz√

1 + 2C2gz=

1

C2g

√

1 + 2C2gz, z =1

2C2g(C4g2λ2 − 1) (238)

Since we want z(λ1) = 0 we see we should choose λ1 = −1/(gC2). z reaches a minimum of−1/(2C2g) at λ = 0 and then increases to 0 at λ = 1/(C2g). We also need x, obtained from

dx

dλ= C

√

−2gz =√

1 − C4g2λ2 (239)

It is convenient to put λ = −(1/C2g) cos θ so dλ/dθ = (1/C2g) sin θ so that

dx

dθ=

1

C2gsin2 θ =

1

2C2g(1 − cos 2θ)

x(θ) =1

2C2g(θ − (1/2) sin 2θ) =

1

4C2g(2θ − sin 2θ) ≡ a(2θ − sin 2θ),

z(θ) =1

2C2g(C4g2λ2 − 1) = − 1

2C2gsin2 θ = − 1

4C2g(1 − cos 2θ) ≡ −a(1 − cos 2θ)

We have chosen the release point to be x = z = 0 or θ = 0. The constant a = 1/(4C2g)must be chosen so that the curve passes through the final point x2, z2.

An interesting feature of the cycloid roller coaster track is that wherever the car is releasedfrom rest on the track, the period of oscillations about the midpoint of the track is the same.This is the subject of the 3 star problem 6.25, for which we sketch the solution. Let us writethe cycloid solution parametrically as follows

x(θ) = a(θ − sin θ), y(θ) = −a(1 − cos θ)) (240)

then the lowest point is θ = π. Then

ds = adθ√

(1 − cos θ)2 + sin2 θ = adθ√

2 − 2 cos θ (241)

and the speed v =√

2ga(cos θ0 − cos θ) where θ0 marks the point the cart is released fromrest. Then a quarter period is the time the cart takes to go from θ0 to π:

T

4=

√

a

g

∫ π

θ0

dθ

√

1 − cos θ

cos θ0 − cos θ=

√

a

g

∫ u0

−1

du√1 − u2

√

1 − u

u0 − u

=

√

a

g

∫ u0

−1

du1

√

(1 + u)(u0 − u)=

√

a

g

∫ 1

0

du1

√

u(1 − u)= π

√

a

g(242)

So the period is just the small oscillation period of a pendulum of length 2a, but for any sizeof amplitude!


8 Hamilton’s Principle of Least (Stationary) Action

8.1 Generalized coordinates

Cartesian coordinates are just fine for describing particles that can move unconstrainedthroughout space. But when the motion is constrained in some way, another choice ofcoordinates may be preferable. As a simple example suppose a particle is constrained tomove in a circle in the xy-plane. Then we have the constraints

z(t) = 0, x2(t) + y2(t) = R2 (243)

We could solve the constraints to eliminate the coordinate y(t) = ±√

R2 − x2(t), but thesign ambiguity is a nuisance. But in passing to polar coordinates x = ρ cos ϕ, y = ρ sin ϕ, wesee that the constraint is simply ρ = R, and ϕ gives a perfectly natural and unambiguousdescription of the particle’s location. Thus in this situation it would be nice to use ϕ andϕ as coordinate and velocity. It is standard to use qk and qk to denote such generalizedcoordinates. There is no need to commit to a particular choice of coordinates in advance.For example a system of N particles can be described by 3N Cartesian coordinates. If thereare k constraints, we can choose s = 3N −k independent generalized coordinates in any waythat is convenient.

8.2 The Action and Hamilton’s Principle

The action is defined as a time integral

I =

∫ t2

t1

dtL(qk(t), qk(t), t) (244)

where L is called the Lagrangian of the system. For the moment we don’t specify it indetail. It is a single scalar function of the generalized coordinates and their velocities, thatdetermines the equations of motion according to Hamilton’s principle: The trajectory qk(t)of the system which starts at the point q1

k at time t1 and ends up at the point q2k at time t2

is that trajectory which minimizes the action I.This means that if we evaluate I for a trajectory qk(t) + δqk(t) infinitesimally different

from the solution, the change in the action will be of order δq2k. So calculate

∆I =

∫ t2

t1

dt [L(qk(t) + δqk(t), qk(t) + δqk(t), t) − L(qk(t), qk(t), t)]

=

∫ t2

t1

dt∑

l

[

δql∂L

∂ql

+ δql(t)∂L

∂ql

]

+ O(δq2)

=∑

l

δql∂L

∂ql

∣

∣

∣

∣

t2

t1

+

∫ t2

t1

dt∑

l

δql

[

∂L

∂ql

− d

dt

∂L

∂ql

]

+ O(δq2) (245)


Now since the ends of the trajectory are fixed, δqk(t1) = δqk(t2) = 0, so qk(t) will satisfyHamilton’s principle if

d

dt

∂L

∂ql

=∂L

∂ql

, for all l (246)

Without saying so, we have just applied what is known as the calculus of variations!If the qk’s are Cartesian coordinates, this will be the form of Newton’s equations if

∂L

∂ql

= mlql(t),∂L

∂ql

= −∂V

∂ql

(247)

which tells us that the Lagrangian in that case can be taken to be

L =1

2

∑

l

mlq2l − V (q) = T (ql) − V (ql) (248)

where T is the kinetic energy of the system and V is the potential energy. Note carefullythe difference of L from the total energy T + V ! there is an all important sign difference inthe second term.

The Lagrangian is not uniquely determined, because Hamilton’s principle requires thatδq(t1) = δq(t2) = 0. For this reason a different Lagrangian

L′ = L +d

dtf(q(t), t), I ′ = I + f(q(t2), t2) − f(q(t1), t1) (249)

will imply the same equations of motion. In other words, two Lagrangians, that differ by thetotal time derivative of a function of coordinates and time, will imply the same equations ofmotion.

8.3 Changing Coordinates in the Lagrangian

One of the virtues of Hamilton’s principle is that the points of stationary action do notdepend on the coordinate choice. Thus if you change coordinates in the action, Lagrange’sequations in the new coordinates still determine the stationary points. It is much easierto change coordinates in the Lagrangian than in the equations of motion! As an examplesuppose we want to use spherical polar coordinates

r = r(cos ϕ sin θ, sin ϕ sin θ, cos θ)

r = r(cos ϕ sin θ, sin ϕ sin θ, cos θ) + r sin θϕ(− sin ϕ, cos ϕ, 0)

+rθ(cos ϕ cos θ, sin ϕ cos θ,− sin θ)

= rr + r sin θϕϕ + rθθ (250)

r2 = r2 + r2θ2 + r2 sin θ2ϕ2 (251)

L =m

2(r2 + r2θ2 + r2 sin θ2ϕ2) − U(r, θ, ϕ) (252)


Notice that if U is independent of ϕ, Lagrange’s equations for ϕ simplify to the conservationof ∂L/∂ϕ = mr2 sin2 θϕ which is just Lz the z-component of angular momentum.

For motion in the xy plane, we can simply put θ = π/2 and the coordinate change is justto polar coordinates r, ϕ:

L =m

2(r2 + r2ϕ2) − U(r, ϕ). (253)

We calculate

∂L

∂r= mr,

∂L

∂r= mrϕ2 − ∂U

∂r, mr = mrϕ2 − ∂U

∂r∂L

∂ϕ= mr2ϕ,

∂L

∂ϕ= −∂U

∂ϕ,

d

dt(mr2ϕ) = −∂U

∂ϕ= rFϕ (254)

It is important to realize that when one changes from Cartesian to other coordinates, thekinetic energy in the new coordinate system can acquire coordinate dependence in additionto velocity dependence, as spherical and polar coordinates demonstrate. Thus in a generalcoordinate system we have T (q, q) but still U(q, t). To incorporate magnetic forces, we willhave to allow velocities to appear in a part of the Lagrangian that is neither kinetic norpotential energy. In other words there will be systems for which the Lagrangian is not of theform T − U !

8.4 The energy from the Lagrangian

We are all familiar with the conservation of energy by Newton’s equations when the forces areconservative. In the Lagrange formulation we can generally identify an energy conservationlaw when the Lagrangian has no explicit time dependence. Consider the Hamiltonian definedby

H ≡∑

i

qi∂L

∂qi

− L (255)

dH

dt=

∑

i

qi∂L

∂qi

+∑

i

qid

dt

∂L

∂qi

−∑

i

qi∂L

∂qi

−∑

i

qi∂L

∂qi

− ∂L

∂t

=∑

i

qi

[

d

dt

∂L

∂qi

− ∂L

∂qi

]

− ∂L

∂t= −∂L

∂t(256)

by Lagrange’s equations. Thus H is conserved provided ∂L/∂t = 0. For standard Newtoniansystems where T is quadratic in the velocities and L = T − V

∑

i

qi∂L

∂qi

= 2T (257)

so H = T + V as we expect.


8.5 The simple pendulum

As a familiar example of a problem with constraints consider the simple frictionless pendulumwith massless rod of length l, swinging in the xz-plane with the pivot at the origin ofcoordinates. Let ϕ be the angle from the vertical. Then x = l sin ϕ, z = −l cos ϕ), andT = (m/2)(x2 + y2) = (ml2/2)ϕ2 and V = −mgl cos ϕ. Hence

L =ml2

2ϕ2 + mgl cos ϕ (258)

and Lagrange’s equation gives the familiar ϕ = −(g/l) sin ϕ. Here we have solved theconstraint x2 + z2 = l2 by going to polar coordinates and setting ρ = l.

Let’s consider the same problem in terms of Cartesian coordinates. Then the uncon-strained Lagrangian is

L =m

2(x2 + z2) − mgz (259)

8.6 Constraints in general

When we consider constraints on a system’s motion, a useful first step is to find the La-grangian for the unconstrained system. This might be a system with several particles withpositions rk(t) moving in a potential U(r1, r2, · · · , rN , t). In this case we immediately knowthe unconstrained Lagrangian L = T − U , and there are 3N degrees of freedom. A typicalconstraint would be to confine the motion to a surface in space. For example, a sphericalpendulum can swing in any direction, with the bob confined to the surface of a sphere. Todescribed the constrained motion it is useful to pick convenient coordinates of the constraintsurface. These generalized coordinates are generically called qk(t) with k = 1, · · ·n < 3N .When the positions rk satisfy the constraints they can be expressed as functions of the qk’s:

rk(q1, · · · qn, t), rk =∑

l

ql∂rk

∂ql

+∂rk

∂t(260)

Plugging these relations into the unconstrained Lagrangian gives the Lagrangian for theconstrained motion L(q, q, t).

As an example consider the spherical pendulum. Introducing spherical coordinates, theposition of the bob can be expressed

r(t) = R(cos ϕ sin θ, sin ϕ sin θ, cos θ) (261)

r(t) = R sin θϕ(− sin ϕ, cos ϕ, 0) + Rθ(cos ϕ cos θ, sin ϕ cos θ,− sin θ) (262)

L =m

2R2 sin2 θϕ2 +

m

2R2θ2 − mgR cos θ (263)

The constrained equations of motion are just the E-L equations:

∂L

∂ϕ= mR2 sin2 θϕ,

∂L

∂θ= mR2θ (264)

∂L

∂ϕ= 0,

∂L

∂θ= mR2 sin θ cos θϕ2 + mgR sin θ (265)


Since the Lagrangian is independent of ϕ, mR2 sin2 θϕ = M is a constant and then

θ = sin θ cos θϕ2 +g

Rsin θ =

M2

m2R4

cos θ

sin3 θ+

g

Rsin θ (266)

this gives a rather complicated equation of motion for M 6= 0! However, one can use energyconservation, as usual, to express t in terms of an integral over θ:

H =m

2R2 sin2 θϕ2 +

m

2R2θ2 + mgR cos θ

=M2

2mR2 sin2 θ+

m

2R2θ2 + mgR cos θ = E (267)

dt

dθ= R

√

m

2

1√

E − mgR cos θ − M2/(2mR2 sin2 θ)(268)

Or consider the double pendulum swinging in a fixed plane.

r1 = l1(sin θ1, cos θ1), r2 = r1 + l2(sin θ2, cos θ2) (269)

or the pendulum on an accelerating cart:

r = (l sin θ + at2/2, l cos θ) (270)

All of these constraints are so-called holonomic constraints, in which the number of coor-dinates needed to describe the system is equal to the number of degrees of freedom: non-holonomic constraints are beyond the scope of this course. (Example: rubber ball rollingwithout slipping on a plane).

Although we won’t be considering non-holonomic constraints further, there is a method ofhandling constraints that avoids eliminating coordinates explicitly, and is a preferred methodfor dealing with non-holonomic constraints.

As an example of this alternative treatment of constraints, let’s return to the simplependulum where we have to impose the holonomic constraint x2 + z2 = l2. The method ofLagrange multipliers adds a term λ(t)(x2 + z2 − l2) to the Lagrangian:

L → m

2(x2 + z2) − mgz + λ(t)(x2 + z2 − l2) (271)

We now regard λ as a generalized coordinate. Since λ doesn’t appear in the Lagrangian, thee.o.m. for λ is just

∂L

∂λ= x2 + y2 − l2 = 0 (272)

which is seen to be precisely the constraint we wish to impose. The e.o.m’s for x, z nowinvolve λ:

mx = 2λx, mz = −mg + 2λz (273)


From this we see that the force exerted by the constraint is F c = 2λ(x, 0, z) ≡ 2λρ. Passingto polar coordinates (x, 0, z) = l(sin ϕ, 0,− cos ϕ) at this point, the e.o.m’s become

ml(ϕ cos ϕ − ϕ2 sin ϕ) = 2λl sin ϕ, 2λ = m(ϕ cot ϕ − ϕ2)

ml(ϕ sin ϕ + ϕ2 cos ϕ) = −mg − 2λl cos ϕ = −mg − ml(ϕ cot ϕ − ϕ2) cos ϕ

ϕ = −g

lsin ϕ, 2λ = −m

(g

lcos ϕ + ϕ2

)

(274)

F c = −m(

g cos ϕ + lϕ2)

(sin ϕ, 0,− cos ϕ) (275)

For example, at the bottom ϕ = 0, F c = m(g + lϕ2)z to compensate gravity and match m×the centripetal acceleration. Notice that if the rod is replaced by a rope, it can only pull onthe particle, which means that it forces the constraint only when λ < 0. This is always trueif ϕ < π/2. But if ϕ > π/2 the rope will only do its job if ϕ2 > −(g/l) cos ϕ!

8.7 Examples

Let’s go through some of the examples in the textbook.

Particle Moving on a Cylinder. In this example the force law is taken to be F = −kr,corresponding to potential energy U = kr2/2 = k(z2 + R2)/2, where the cylinder axiscoincides with the z-axis and R is the radius of the cylinder. The kinetic energy of theparticle is (m/2)(z2 + R2ϕ2). thus the Lagrangian is

L =m

2(z2 + R2ϕ2) − k

2(z2 + R2) (276)

∂L

∂ϕ= mR2ϕ,

∂L

∂ϕ= 0,

∂L

∂z= mz,

∂L

∂z= −kz (277)

Clearly z undergoes simple harmonic motion with ω =√

k/m and ϕ is a constant.

Block sliding on a wedge free to move on a horizontal surface: Let q1 be thedisplacement along the wedge of the block from the top of the wedge, and q2 be the horizontaldisplacement of the left edge of the wedge. The kinetic energy of the wedge is just m2q

22/2.

The Cartesian coordinates of the block are (q2 + q1 cos α, 0, q1 sin α) where α is the angle thewedge makes with the horizontal. the block’s kinetic energy is therefore m1[(q2 + q1 cos α)2 +q21 sin2 α]/2. The block’s potential energy is −mgq1 sin α. Putting all this together

L =m2

2q22 +

m1

2(q2

2 + q21 + 2q2q1 cos α) + m1gq1 sin α (278)

∂L

∂q1

= m1q1 + m1q2 cos α,∂L

∂q2

= (m1 + m2)q2 + m1q1 cos α

∂L

∂q1

= m1g sin α,∂L

∂q2

= 0

q2 = −m1 cos α

m1 + m2

q1, m1q1 + m1q2 cos α = m1g sin α

q1 =g sin α

1 − m1 cos2 α/(m1 + m2)(279)


Note that this goes to the answer for a fixed wedge when m2/m1 → ∞. It also goes to freefall when α → π/2.

Bead on a spinning frictionless circular hoop The bead’s position vector is

r = R(cos ωt sin θ, sin ωt sin θ,− cos θ)

r = R sin θω(− sin ωt, cos ωt, 0) + Rθ(cos ωt cos θ, sin ωt cos θ, sin θ)

L =m

2(ω2R2 sin2 θ + R2θ2) + mgR cos θ (280)

Notice that in spite of the forced rotation, L does not depend explicitly on the time. Thereforethe Hamiltonian is a constant of the motion:

H = θ(mR2θ) − L =m

2(R2θ2 − ω2R2 sin2 θ) − mgR cos θ (281)

This is not the total energy of the bead which is

T + U =m

2(ω2R2 sin2 θ + R2θ2) − mgR cos θ = H + mω2R2 sin2 θ (282)

which is not conserved. The E-L equations are:

mR2θ = −mgR sin θ + mω2R2 sin θ cos θ = mR2 sin θ(

ω2 cos θ − g

R

)

(283)

Static solutions are possible for θ = 0, π and for cos θ = g/(Rω2) if ω >√

g/R. θ = π is

always unstable; θ = 0 is stable for ω <√

g/R, but becomes unstable for ω >√

g/R. Inthe latter case, put θ0 = arccos(g/Rω2) and expand

cos θ =g

Rω2− (θ − θ0) sin θ0 + O(θ − θ0)

2 (284)

so the right side behaves near θ = θ0 as

−mR2ω2 sin2 θ0(θ − θ0) = −mR2

(

ω2 − g2

R2ω2

)

(θ − θ0) (285)

Thus the solution θ = θ0 is stable, and harmonic small oscillations about it have angularfrequency ω0 =

√

ω2 − g2/(R2ω2).A nice way to think about the different behaviors of this system is the effective potential

energy

Ueff(θ) = −mgR cos θ − 1

2mω2R2 sin2 θ (286)

plotted as a function of θ in the following figure (a = ω2R/g):



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